67486 + Interview Questions in Secondary School in Physics

1.	8: A balloon is inflated in a cold room. When the room becomes much warmer theballoon becomeslarger. How does the behaviour of the air molecules in the balloonexplain this?A) The molecules become largerB) The molecules evaporateC) The molecules move more quickD) The molecules repel each other.
Answer» Answer: C ) the molecules move more quick. Explanation: Due to increase in surrounding heat, the kinetic energy of air molecules increase and they start MOVING more quickly. The interparticular collisons and collison with balloon's INNER wall exerts PRESSURE on the balloon from inside, thus increasing the size of balloon. Thanks, have a GREAT DAY ahead!

1.

8: A balloon is inflated in a cold room. When the room becomes much warmer theballoon becomeslarger. How does the behaviour of the air molecules in the balloonexplain this?A) The molecules become largerB) The molecules evaporateC) The molecules move more quickD) The molecules repel each other.

Answer»

Answer:

C ) the molecules move more quick.

Explanation:

Due to increase in surrounding heat, the kinetic energy of air molecules increase and they start MOVING more quickly.

The interparticular collisons and collison with balloon's INNER wall exerts PRESSURE on the balloon from inside, thus increasing the size of balloon.

Thanks, have a GREAT DAY ahead!

Discussion

2.	(d) None133. Two bodies 'A' and 'B' have velocities in theratio of 2 : 1 mass of body A is half of the mass ofbody 'B' the kinetic energies of 'A' and 'B' are inthe ratio of
Answer» Given : • Two BODIES 'A' and 'B' have velocities in the ratio of 2 : 1. • Mass of body A is half of the mass of body 'B'. To calculate : • The ratio of kinetic energies of A and B. Calculation : As velocities of 'A' and 'B' are in the ratio of 2 : 1. So, Let , • Velocity of body A = 2v • Velocity of body B = 1v or v Also, let us ASSUME , • Mass of body B = m So, as per the given question mass of body A is half of the mass of body 'B'. $\longrightarrow$ Mass of Body A = $\sf { \dfrac{1}{2} m }$ Now, let's calculate the ratio of the kinetic energies of 'A' and 'B' . $\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$ Kinetic Energy of Body A : We KNOW that, $\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}} \\$ • K.E = Kinetic Energy • m = mass • v = velocity Here, ✰ Mass of A = $\sf { \dfrac{1}{2}m}$ ✰ Velocity of A = 2v Substituting VALUES, $\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{2} \times \dfrac{1}{2}m \times {(2v)}^{2} }$ $\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{\cancel{4}}m \times {\cancel{4}v}^{2} }$ $\longrightarrow \sf {K.E_{(Body \: A)} = 1m \times {v}^{2} }$ $\longrightarrow \boxed{\sf {K.E_{(Body \: A)} = m{v}^{2}} }$ Kinetic Energy of Body B : We know that, $\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}} \\$ • K.E = Kinetic Energy • m = mass • v = velocity Here, ✰ Mass of B = m ✰ Velocity of B = v Substituting values, $\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {(v)}^{2} }$ $\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {v}^{2} }$ $\longrightarrow \boxed{ \sf {K.E_{(Body \: B)} = \dfrac{1}{2}m{v}^{2}} }$ Calculating Ratio : $\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$ $\longrightarrow \sf { Ratio = \dfrac{ \cancel{m{v}^{2}} }{ \cfrac{1}{2} \cancel{m {v}^{2}}} }$ $\longrightarrow \sf { Ratio = \dfrac{1 }{ \cfrac{1}{2}} }$ $\longrightarrow \sf { Ratio = 1 \times \dfrac{2}{1}}$ $\longrightarrow \sf { Ratio = \dfrac{2}{1}}$ $\longrightarrow \boxed {\pmb{ \rm \red { Ratio = 2:1}}}$ Therefore, the kinetic energies of 'A' and 'B' are in the ratio of 2 : 1.

2.

(d) None133. Two bodies 'A' and 'B' have velocities in theratio of 2 : 1 mass of body A is half of the mass ofbody 'B' the kinetic energies of 'A' and 'B' are inthe ratio of

Answer»

Given :

• Two BODIES 'A' and 'B' have velocities in the

ratio of 2 : 1.

• Mass of body A is half of the mass of body 'B'.

To calculate :

• The ratio of kinetic energies of A and B.

Calculation :

As velocities of 'A' and 'B' are in the ratio of 2 : 1. So,

Let ,

• Velocity of body A = 2v

• Velocity of body B = 1v or v

Also, let us ASSUME ,

• Mass of body B = m

So, as per the given question mass of body A is half of the mass of body 'B'.

$\longrightarrow$ Mass of Body A = $\sf { \dfrac{1}{2} m }$

Now, let's calculate the ratio of the kinetic energies of 'A' and 'B' .

$\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$

Kinetic Energy of Body A :

We KNOW that,

$\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}} \\$

• K.E = Kinetic Energy

• m = mass

• v = velocity

Here,

✰ Mass of A = $\sf { \dfrac{1}{2}m}$

✰ Velocity of A = 2v

Substituting VALUES,

$\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{2} \times \dfrac{1}{2}m \times {(2v)}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{\cancel{4}}m \times {\cancel{4}v}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: A)} = 1m \times {v}^{2} }$

$\longrightarrow \boxed{\sf {K.E_{(Body \: A)} = m{v}^{2}} }$

Kinetic Energy of Body B :

We know that,

$\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}} \\$

• K.E = Kinetic Energy

• m = mass

• v = velocity

Here,

✰ Mass of B = m

✰ Velocity of B = v

Substituting values,

$\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {(v)}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {v}^{2} }$

$\longrightarrow \boxed{ \sf {K.E_{(Body \: B)} = \dfrac{1}{2}m{v}^{2}} }$

Calculating Ratio :

$\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$

$\longrightarrow \sf { Ratio = \dfrac{ \cancel{m{v}^{2}} }{ \cfrac{1}{2} \cancel{m {v}^{2}}} }$

$\longrightarrow \sf { Ratio = \dfrac{1 }{ \cfrac{1}{2}} }$

$\longrightarrow \sf { Ratio = 1 \times \dfrac{2}{1}}$

$\longrightarrow \sf { Ratio = \dfrac{2}{1}}$

$\longrightarrow \boxed {\pmb{ \rm \red { Ratio = 2:1}}}$

Therefore, the kinetic energies of 'A' and 'B' are in

the ratio of 2 : 1.

Discussion

3.	Match the column - with column-ll and mark the appropriate choiceColumn-1Column-11(A) To measure mass(P) Doppler's effect of light(B) To measure time(Q Law of gravitation and spectrometer(C) To measure distance(R) Caesium clock(D) To measure speed of star (S) RADAR and SONAROA-Q, B-R, C-S, D-Po A- PQ, B- QR, C- PQR, D-So A-Q, B-R, C-P. D-SKi
Answer» 10 11G 1H 3H 3h4v5h make me BRAINLEAST

3.

Match the column - with column-ll and mark the appropriate choiceColumn-1Column-11(A) To measure mass(P) Doppler's effect of light(B) To measure time(Q Law of gravitation and spectrometer(C) To measure distance(R) Caesium clock(D) To measure speed of star (S) RADAR and SONAROA-Q, B-R, C-S, D-Po A- PQ, B- QR, C- PQR, D-So A-Q, B-R, C-P. D-SKi

Answer»

10 11G 1H 3H 3h4v5h

make me BRAINLEAST

Discussion

4.	An object is dropped from a balloon rising up with a velocity 2 ms-1. Find the velocity of the object after 2 secondsof its release. (take g = 10 ms-2)(1) 9 ms-1(2) 18 ms(3) 27 ms-1(4) 36 ms-1
Answer» Answer: Here, u = 2 ms-1, t = 2S, a = 10 ms-2 Using v = u + at We get v = 2 + 10 × 2 => v = 22 ms-1 Thus, VELOCITY of the object after 2 SECONDS will be 22 ms-1

4.

An object is dropped from a balloon rising up with a velocity 2 ms-1. Find the velocity of the object after 2 secondsof its release. (take g = 10 ms-2)(1) 9 ms-1(2) 18 ms(3) 27 ms-1(4) 36 ms-1

Answer»

Answer:

Here, u = 2 ms-1, t = 2S, a = 10 ms-2

Using

v = u + at

We get

v = 2 + 10 × 2

=> v = 22 ms-1

Thus, VELOCITY of the object after 2 SECONDS will be 22 ms-1

Discussion

5.	Define the continuous and composite magnetic circuits with examples. Prove that for thegiven magnetic circuit (see Figure) total mmf is Hal1+Hz]2+H3l3+Hsls.la,
Answer» Answer: This lecture NOTE is BASED on the textbook # 1. Electric Machinery - A.E. Fitzgerald, CHARLES KINGSLEY, Jr., Stephen D. Umans- 6th edition- Mc Graw Hill series in Electrical Engineering. Power and Energy The objective of this course is to STUDY the devices used in the interconversion of electric and mechanical energy, with emphasis placed on electromagnetic rotating machinery. The transformer, although not an electromechanical-energy-conversion device, is an important component of the overall energy-conversion process. Practically all transformers and electric machinery use ferro-magnetic material for shaping and directing the magnetic fields that acts as the medium for transferring and converting energy. Permanent-magnet materials are also widely used. The ability to analyze and describe systems containing magnetic materials is essential for designing and understanding electromechanical-energy-conversion devices. Explanation: mark me as brainlist

5.

Define the continuous and composite magnetic circuits with examples. Prove that for thegiven magnetic circuit (see Figure) total mmf is Hal1+Hz]2+H3l3+Hsls.la,

Answer»

Answer:

This lecture NOTE is BASED on the textbook # 1. Electric Machinery - A.E. Fitzgerald, CHARLES KINGSLEY, Jr., Stephen D. Umans- 6th edition- Mc Graw Hill series in Electrical Engineering. Power and Energy

The objective of this course is to STUDY the devices used in the interconversion of electric and mechanical energy, with emphasis placed on electromagnetic rotating machinery.

The transformer, although not an electromechanical-energy-conversion device, is an important component of the overall energy-conversion process.

Practically all transformers and electric machinery use ferro-magnetic material for shaping and directing the magnetic fields that acts as the medium for transferring and converting energy. Permanent-magnet materials are also widely used.

The ability to analyze and describe systems containing magnetic materials is essential for designing and understanding electromechanical-energy-conversion devices.

Explanation:

mark me as brainlist

Discussion

6.	A crown weighs 35N in air and 32N inwater. Determine its density
Answer» Answer:11666.66666666667Kg/m^3 OR 11666.6Kg/m^3 Explanation: (σ=density) σρ=Weight(AIR)/W(air)−W(LIQUID) σ/1000=35/35–32 σ=35x1000/35–32 σ=35000/3 σ=11666.66666666667Kg/m^3 (approx=11666.6Kg/m^3)

6.

A crown weighs 35N in air and 32N inwater. Determine its density

Answer»

Answer:11666.66666666667Kg/m^3 OR 11666.6Kg/m^3

Explanation: (σ=density)

σρ=Weight(AIR)/W(air)−W(LIQUID)

σ/1000=35/35–32

σ=35x1000/35–32

σ=35000/3

σ=11666.66666666667Kg/m^3 (approx=11666.6Kg/m^3)

Discussion

7.	Draw ray diagram for the following and label them. Also, write the measures of the angle of incidence and reflection.A ray striking a plane mirror along the normal. A ray striking a plane surface making an angle of 300 with it. Two plane mirrors with their reflecting surfaces perpendicular to each other and a ray of light striking one of the mirrors making an angle of 400 with it.
Answer» ANSWER: Tok vi don naneun RULES woahhahhao opshi do nareul rules wiahhahhao opshi GE LIGHTS out tonight

7.

Draw ray diagram for the following and label them. Also, write the measures of the angle of incidence and reflection.A ray striking a plane mirror along the normal. A ray striking a plane surface making an angle of 300 with it. Two plane mirrors with their reflecting surfaces perpendicular to each other and a ray of light striking one of the mirrors making an angle of 400 with it.

Answer»

ANSWER:

Tok vi don naneun RULES

woahhahhao

opshi do nareul rules

wiahhahhao

opshi GE LIGHTS out tonight

Discussion

8.	Define speed and velocity. Write their Sl units. A body is moving with a velocityof 15 m/s. If the motion is uniform, what will be the velocity after 10 s?
Answer» Answer: Can any ONE TELL me how to ADD questions in BRAINLY

8.

Define speed and velocity. Write their Sl units. A body is moving with a velocityof 15 m/s. If the motion is uniform, what will be the velocity after 10 s?

Answer»

Answer:

Can any ONE TELL me how to ADD questions in BRAINLY

Discussion

9.	Tera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTu jo mere sang chalne lageTo meri raahein dhadakne lageDekhun jo na ik pal main tumheinToh meri baahein tadapne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reHaathon se laqeerein yehi kehti haiKi zindagi jo hai meriTujhi mein ab rehti haiLabon pe likhi hai mere dil ki khwaahishLafzon mein kaise main bataaunIkk tujhko hi paane ki khaatirSabse judaa main ho jaaunKal tak maine jo bhi khwaab the dekheTujhme woh dikhne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reSaanson ke kinaare bade tanha thheTu aa ke inhe chhu le basYehi toh mere armaan thheSaari duniya se mujhe kya lena haiBas tujhko hi pehchaanuMujhko na meri ab khabar ho koiTujhse hi khudko main jaanuRaatein nahi kat’ti bechain se hokeDin bhi guzarne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya re. write by - anish my friend
Answer» Photosynthesis, the process by which green PLANTS and certain other organisms transform light energy into CHEMICAL energy. During photosynthesis in green plants, light energy is captured and used to CONVERT water, CARBON dioxide, and minerals into oxygen and energy-rich ORGANIC compounds. Achha hai..✔

9.

Tera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTu jo mere sang chalne lageTo meri raahein dhadakne lageDekhun jo na ik pal main tumheinToh meri baahein tadapne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reHaathon se laqeerein yehi kehti haiKi zindagi jo hai meriTujhi mein ab rehti haiLabon pe likhi hai mere dil ki khwaahishLafzon mein kaise main bataaunIkk tujhko hi paane ki khaatirSabse judaa main ho jaaunKal tak maine jo bhi khwaab the dekheTujhme woh dikhne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya reSaanson ke kinaare bade tanha thheTu aa ke inhe chhu le basYehi toh mere armaan thheSaari duniya se mujhe kya lena haiBas tujhko hi pehchaanuMujhko na meri ab khabar ho koiTujhse hi khudko main jaanuRaatein nahi kat’ti bechain se hokeDin bhi guzarne lageIshq jo zara sa tha woh badh gaya reTera fitoor jab se chadh gaya reTera fitoor jab se chadh gaya re. write by - anish my friend

Answer»

Photosynthesis, the process by which green PLANTS and certain other organisms transform light energy into CHEMICAL energy. During photosynthesis in green plants, light energy is captured and used to CONVERT water, CARBON dioxide, and minerals into oxygen and energy-rich ORGANIC compounds.

Achha hai..✔

Discussion

10.	A person standing between the two vertical cliffs produces sound twosuccessive echoes heard at 4s and 6s. Calculate the distance between cliffs? +
Answer» ANSWER: I HOPE it will be HELP FULL for you

10.

A person standing between the two vertical cliffs produces sound twosuccessive echoes heard at 4s and 6s. Calculate the distance between cliffs? +

Answer»

ANSWER:

I HOPE it will be HELP FULL for you

Discussion

11.	The following types of solid contain molecules as constituent particles.(a) molecular solid ,(b) ionic solids, (c) metallic solid (d) covalent network solids.
Answer» ANSWER: Here is your answer Explanation: B. IONIC SOLIDS

11.

The following types of solid contain molecules as constituent particles.(a) molecular solid ,(b) ionic solids, (c) metallic solid (d) covalent network solids.

Answer»

ANSWER:

Here is your answer

Explanation:

B. IONIC SOLIDS

Discussion

12.	All the Shorinons of each chapter are ale inat...evropy i zeloose when 50 gm ofmers isof How muchtoexicono completed teacher
Answer» Answer: Middle EAST respiratory syndrome (MERS) is a respiratory DISEASE CAUSED by a CORONAVIRUS (MERS- CoV). The disease was first reported in Saudi ARABIA in 2012.

12.

All the Shorinons of each chapter are ale inat...evropy i zeloose when 50 gm ofmers isof How muchtoexicono completed teacher

Answer»

Answer:

Middle EAST respiratory syndrome (MERS) is a respiratory DISEASE CAUSED by a CORONAVIRUS (MERS- CoV). The disease was first reported in Saudi ARABIA in 2012.

Discussion

13.	Question No. 3If work =x^2/beta e alpha ktWhere, K = Boltzmann constant, T = temp, then find dimension of beta/alpha :-O ML2O M²LO M²L214ОМL3
Answer» ANSWER: please mark this brainliest please again I FOLLOW and like you

13.

Question No. 3If work =x^2/beta e alpha ktWhere, K = Boltzmann constant, T = temp, then find dimension of beta/alpha :-O ML2O M²LO M²L214ОМL3

Answer»

ANSWER:

please mark this brainliest please again I FOLLOW and like you

Discussion

14.	In Two port Network,condition of symmetry forABCD parameters is *a) B = Cb)A = Dc) B =Dd) A=C
Answer» In physics, a FORCE is any interaction that, when unopposed, will change the MOTION of an object. A force can cause an object with MASS to change its velocity, i.e., to accelerate. Force can also be described intuitively as a PUSH or a pull. A force has both magnitude and direction, making it a VECTOR quantity.

Discussion

15.	UIUuo0=tan 1Q. 14. A small body of mass m is tied to a string and revolved in avertical circle of radius r. If the tension in the string at the highestpoint is mg, what is its speed there?
Answer» Answer: 8 mg Explanation: LET lowest point val. = v work energy principle wall=0K w G +w Y =K f −K i [∵w p = work done by tension = 0] mg(2r)= 2 1 m( 3rg ) 2 − 2 1 mv 2 2mgr= 2 3 mgr− 2 1 mv 2 mv 2 =3mgr+4mgr ∴v= 7gr ∴F =T−mg r m( 7gr ) 2 =T−mg ⇒T=7mg+mg=8mg Hope it helps you PLS mark me as Brainliest

15.

UIUuo0=tan 1Q. 14. A small body of mass m is tied to a string and revolved in avertical circle of radius r. If the tension in the string at the highestpoint is mg, what is its speed there?

Answer»

Answer:

8 mg

Explanation:

LET lowest point val. = v

work energy principle

wall=0K

w

G

+w

Y

=K

f

−K

i

[∵w

p

= work done by tension = 0]

mg(2r)=

2

1

m(

3rg

)

2

−

2

1

mv

2

2mgr=

2

3

mgr−

2

1

mv

2

mv

2

=3mgr+4mgr

∴v=

7gr

∴F

=T−mg

r

m(

7gr

)

2

=T−mg

⇒T=7mg+mg=8mg

Hope it helps you

PLS mark me as Brainliest

Discussion

16.	Next four meqs based on the comprehension given below. Choose the correct option.A person is interested to find experimentally a relation between range of the ball and velocity of projection of ball Themesthe horizontal distance upto which the ball goes on ground after projection with some velocity on ground let the ground bewhere the person is performing the experiment in each projection the person uses the same ball and project it at the sameProjectionProjectionspeed(m/s)Range(m)151016133620he table show the experimental detail. From above observation the person concludes that the range R depend from the ineed v according to the equation; R-CV-t, where C is a constant and n is another constant. Answer the following:2. On the basis of above observation what is the value of n?a) 2.5b) 3.0c) 3.5d) 4.00. The person performs another experiment in which the ball is projected at a speed of 4 ms its range is approxima) 2.12m b) 3.45 mc) 2.56 m d) 4.02 mir is the angle of projection of the ball with velocity v from the horizontal ground the speed of the ball at the hisse trajectory isa) 0 b) y sinoc) v cose d) v2From the observation given in the table the person thinks that C depends onhich the hall was projectedii) the mass of the ballITAKIMI A1
Answer» Answer: २४६७४५८२७२७+६२७२६३६४२_२६२८२६-८१९७२३६+₹(-१८८२६३३८२७-#:#!#&३८३९२७६३२८९२३७३

16.

Next four meqs based on the comprehension given below. Choose the correct option.A person is interested to find experimentally a relation between range of the ball and velocity of projection of ball Themesthe horizontal distance upto which the ball goes on ground after projection with some velocity on ground let the ground bewhere the person is performing the experiment in each projection the person uses the same ball and project it at the sameProjectionProjectionspeed(m/s)Range(m)151016133620he table show the experimental detail. From above observation the person concludes that the range R depend from the ineed v according to the equation; R-CV-t, where C is a constant and n is another constant. Answer the following:2. On the basis of above observation what is the value of n?a) 2.5b) 3.0c) 3.5d) 4.00. The person performs another experiment in which the ball is projected at a speed of 4 ms its range is approxima) 2.12m b) 3.45 mc) 2.56 m d) 4.02 mir is the angle of projection of the ball with velocity v from the horizontal ground the speed of the ball at the hisse trajectory isa) 0 b) y sinoc) v cose d) v2From the observation given in the table the person thinks that C depends onhich the hall was projectedii) the mass of the ballITAKIMI A1

Answer»

Answer:

२४६७४५८२७२७+६२७२६३६४२_२६२८२६-८१९७२३६+₹(-१८८२६३३८२७-#:#!#&३८३९२७६३२८९२३७३

Discussion

17.	Question 2 :The three-phase winding shown in figure havebeen connected in theClick on image to enlargeА в сDelta configurationDouble series configurationParallel configurationStar configuration
Answer» Shiwalayon a WA wmein will SEND me your ADDRESS

Discussion

18.	Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is themagnitude of their resultant?
Answer» HOPE This is will HELP U _______________

18.

Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is themagnitude of their resultant?

Answer»

HOPE This is will HELP U

_______________

Discussion

19.	Question 1:A RLC parallel circuit is.acceptor circuitrejector circuituniversal circuitsender circuit
Answer» Answer: universal CIRCUIT EXPLANATION: hope it helps U!!!!!!!!!!!

19.

Question 1:A RLC parallel circuit is.acceptor circuitrejector circuituniversal circuitsender circuit

Answer»

Answer:

universal CIRCUIT

EXPLANATION:

hope it helps U!!!!!!!!!!!

Discussion

20.	A satellite of mass m is in circular orbit around the Earth at a distance R from the Earth's center. If R is increased by a factor of 4, then the period of the orbitwill be :(1) unchanged(2) increased by a factor of 4(3) increased by a factor of 8 (4) increased by a factor of 64
Answer» Answer: 13.4 Satellite Orbits and Energy Learning Objectives By the end of this section, you will be able to: Describe the mechanism for circular orbits Find the orbital periods and speeds of satellites Determine whether objects are gravitationally bound The Moon orbits Earth. In turn, Earth and the other PLANETS orbit the Sun. The space directly above our ATMOSPHERE is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit. Circular Orbits As noted at the beginning of this chapter, Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a MERE 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%. Determining the orbital speed and orbital period of a satellite is much EASIER for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases. Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth (Figure 13.12). It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives GmMEr2=mac=mv2orbitr. A drawing shows a satellite orbiting the earth at radius r. The orbit is shown as a blue circle centered on the earth. A RED arrow at the satellite points toward the center of the earth and is labeled F and a green arrow tangent to the orbit is labeled v. Figure 13.12 A satellite of mass m orbiting at radius r from the

20.

A satellite of mass m is in circular orbit around the Earth at a distance R from the Earth's center. If R is increased by a factor of 4, then the period of the orbitwill be :(1) unchanged(2) increased by a factor of 4(3) increased by a factor of 8 (4) increased by a factor of 64

Answer»

Answer:

13.4 Satellite Orbits and Energy

Learning Objectives

By the end of this section, you will be able to:

Describe the mechanism for circular orbits

Find the orbital periods and speeds of satellites

Determine whether objects are gravitationally bound

The Moon orbits Earth. In turn, Earth and the other PLANETS orbit the Sun. The space directly above our ATMOSPHERE is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit.

Circular Orbits

As noted at the beginning of this chapter, Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a MERE 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%.

Determining the orbital speed and orbital period of a satellite is much EASIER for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases.

Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth (Figure 13.12). It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives

GmMEr2=mac=mv2orbitr.

A drawing shows a satellite orbiting the earth at radius r. The orbit is shown as a blue circle centered on the earth. A RED arrow at the satellite points toward the center of the earth and is labeled F and a green arrow tangent to the orbit is labeled v.

Figure 13.12 A satellite of mass m orbiting at radius r from the

Discussion

21.	1. Calculate the velocity of a body of mass 0.3[10 ms2. An electron of mass 9 ~ 10 kg is moving with a linear velocity of 6 * 10 ms Calculate thelinear momentum of electron154 10-Nsm 200 is moving with a velocity of 5 ms. If the velocity of the body changes12.4 Ns]
Answer» ANSWER: okkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkk

21.

1. Calculate the velocity of a body of mass 0.3[10 ms2. An electron of mass 9 ~ 10 kg is moving with a linear velocity of 6 * 10 ms Calculate thelinear momentum of electron154 10-Nsm 200 is moving with a velocity of 5 ms. If the velocity of the body changes12.4 Ns]

Answer»

ANSWER:

okkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkk

Discussion

22.	Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.[10 ms-']2. An electron of mass 9 ~ 10-31 kg is moving with a linear velocity of 6 x 107 msl. Calculate thelinear momentum of electron.[54 x 10-24 Ns]
Answer» ANSWER: PLZ MARK as BRAINLIST PLEASE

22.

Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.[10 ms-']2. An electron of mass 9 ~ 10-31 kg is moving with a linear velocity of 6 x 107 msl. Calculate thelinear momentum of electron.[54 x 10-24 Ns]

Answer»

ANSWER:

PLZ MARK as BRAINLIST PLEASE

Discussion

23.	When monochromatic light with a wavelength of 430 nm irradiates aphotocathode a minimum stopping potential of 0.65 V is required tostop the photoelectric current.a. Determine the photoelectric work function.b. calculate threshold wavelength.
Answer» Answer: the photoelectric WORK function is that of the MINIMUM stopping potential and the second ONE ???

23.

When monochromatic light with a wavelength of 430 nm irradiates aphotocathode a minimum stopping potential of 0.65 V is required tostop the photoelectric current.a. Determine the photoelectric work function.b. calculate threshold wavelength.

Answer»

Answer:

the photoelectric WORK function is that of the MINIMUM stopping potential and the second ONE ???

Discussion

24.	18. In Raman spectrum, if I is thewavelength of incident radiation, thenthe Stoke's lines will have wavelengthequal toO a)Ο b)λ+ ΔλΟ c)λ - ΔλO d) 212
Answer» The ANSWER is MINE Is SAYING to ADD you because

Discussion

25.	The diagram shows four acting on a block What is the resultant force?(a) Zero6 N to right(b)5 N to left(d) 11 N to right
Answer» Answer: HEY MATE !! HERE IS UR ANS ⤵️⤵️ option b 11 Newton to right is ur correct ans ! ... HOPE THIS HELPS U ALMOST ... ... THANK Q ... ꧁꧁ ༒ λKŞU ༒꧂꧂

25.

The diagram shows four acting on a block What is the resultant force?(a) Zero6 N to right(b)5 N to left(d) 11 N to right

Answer»

Answer:

HEY MATE !!

HERE IS UR ANS ⤵️⤵️

option b 11 Newton to right is ur correct ans !

... HOPE THIS HELPS U ALMOST ...

... THANK Q ...

꧁꧁ ༒ λKŞU ༒꧂꧂

Discussion

26.	What is the percentage change in length of 4m iron rod if its temperature changes by 200°C.(a iron = 2 x 10^-5/°C)O 0.2%O 0.4%O 0.6%O 0.5%
Answer» We have to find the percentage change in length of 4m iron rod if its TEMPERATURE changes by 200°C. coefficient of iron is 2 × 10¯⁵/°C. solution : change in length is given by, ∆L = L₀α∆T ⇒fractional change = ∆L/L₀ = α∆T ⇒percentage change = ∆L/L₀ × 100 = 100α∆T here α = 2 × 10¯⁵/°C and ∆T = 200°C so, percentage change in length of iron rod = 100 × 2 × 10¯⁵ × 200 = 4 × 10⁴ × 10¯⁵ = 0.4 % Therefore the correct option is (B) 0.4 % ALSO read similar questions : Why a brown coating is formed on the iron rod when iron rod is kept dipped in copper sulphate solution for sometime? Wha... brainly.in/question/2218504 An iron rod of lenght 30cm is heated through 50 kelvin CALCULATE its increase in lenght (LINEAR expansivity of iron=1.2 ... brainly.in/question/17700596

26.

What is the percentage change in length of 4m iron rod if its temperature changes by 200°C.(a iron = 2 x 10^-5/°C)O 0.2%O 0.4%O 0.6%O 0.5%

Answer»

We have to find the percentage change in length of 4m iron rod if its TEMPERATURE changes by 200°C. coefficient of iron is 2 × 10¯⁵/°C.

solution : change in length is given by, ∆L = L₀α∆T

⇒fractional change = ∆L/L₀ = α∆T

⇒percentage change = ∆L/L₀ × 100 = 100α∆T

here α = 2 × 10¯⁵/°C and ∆T = 200°C

so, percentage change in length of iron rod = 100 × 2 × 10¯⁵ × 200

= 4 × 10⁴ × 10¯⁵ = 0.4 %

Therefore the correct option is (B) 0.4 %

ALSO read similar questions : Why a brown coating is formed on the iron rod when iron rod is kept dipped in copper sulphate solution for sometime? Wha...

brainly.in/question/2218504

An iron rod of lenght 30cm is heated through 50 kelvin CALCULATE its increase in lenght (LINEAR expansivity of iron=1.2 ...

brainly.in/question/17700596

Discussion

27.	If mass M and radius of solid sphere is R, then moment of Inertia aboutdiameter is :(A) 2/3MR^2(B) 2/5MR^2(c) 3/5MR^2(d) 7/5MR^2
Answer» A solid sphere of mass M and radius R having moment of INERTIA I about its diameter is recast into a solid disc ABUT an axis passing the edge and perpendicular to the PLANE REMAINS l. Then R and r are related as. From(i) and (ii), 25MR2=32Mr2⇒r=2√15R.

Discussion

28.	Who was the first scientist to arrange the elements in atomic no??Guys pls follow it's in my following
Answer» Answer: DIMITRI Endeeleve.. YUP sure have a bangtastic day AHEAD:) Saranghae

28.

Who was the first scientist to arrange the elements in atomic no??Guys pls follow it's in my following

Answer»

Answer:

DIMITRI Endeeleve..

YUP sure have a bangtastic day AHEAD:)

Saranghae

Discussion

29.	No part of the energy ever returns to the sun, because passge of energy is concidered to be a ______ transport.1️⃣ one way2️⃣ two way3️⃣ multiway4️⃣ any direction
Answer» ANSWER: ONE Way Explanation:

29.

No part of the energy ever returns to the sun, because passge of energy is concidered to be a ______ transport.1️⃣ one way2️⃣ two way3️⃣ multiway4️⃣ any direction

Answer»

ANSWER:

ONE Way

Explanation:

Discussion

30.	Write relation between coefficient of linear expansion (alpha) and coefficient ofvolume expansion (r) of a solid material.
Answer» Answer: RELATIONSHIP to Linear Thermal EXPANSION Coefficient For isotropic material, and for small expansions, the linear thermal expansion coefficient is ONE third the VOLUMETRIC coefficient. To derive the relationship, let's TAKE a cube of steel that has sides of length L.

30.

Write relation between coefficient of linear expansion (alpha) and coefficient ofvolume expansion (r) of a solid material.

Answer»

Answer:

RELATIONSHIP to Linear Thermal EXPANSION Coefficient

For isotropic material, and for small expansions, the linear thermal expansion coefficient is ONE third the VOLUMETRIC coefficient. To derive the relationship, let's TAKE a cube of steel that has sides of length L.

Discussion

31.	Que.What do you meanby dispersion? Name thedifferent colour of light inthe proper sequence in thespectrum of light?
Answer» Answer: When white light is passed through a GLASS PRISM it splits into its spectrum of colours and this PROCESS of white light splitting into its constituent colours is termed as dispersion. violet, indigo, blue, green, YELLOW, ORANGE and red

31.

Que.What do you meanby dispersion? Name thedifferent colour of light inthe proper sequence in thespectrum of light?

Answer»

Answer:

When white light is passed through a GLASS PRISM it splits into its spectrum of colours and this PROCESS of white light splitting into its constituent colours is termed as dispersion.

violet, indigo, blue, green, YELLOW, ORANGE and red

Discussion

32.	WHAT IS VERB.........ACCOUNT DELETION
Answer»

Discussion

33.	Q] Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked roadAndQ] Show that the angle of banking is independent of mass of vehicle.
Answer» Refer to the attachment first for better understanding! (1) Net force acting on the vehicle, $\to\sf{N cos \theta = mg + f sin \theta } \\$ $\to\sf{N cos \theta- f sin \theta = mg ....(1)} \\$ $\to\sf{N sin \theta + f cos \theta =\dfrac{ mv^<klux>2</klux> }{ r} ....(2)} \\ \\$ Dividing (2) and (1) we get, $\to\sf{\dfrac{N sin \theta + f cos \theta }{N cos \theta - f sin \theta } = \dfrac{v^2}{ rg}}\\ \\$ Now LET's divide the numerator and denominator of the L.H.S by N cos θ, $\to\sf{\dfrac{\dfrac{N sin \theta }{N cos \theta } + \dfrac{f cos \theta }{N cos \theta }}{\dfrac{N cos \theta }{N cos \theta } - \dfrac{f sin \theta }{N cos \theta }} = \dfrac{v^2}{ rg}} \\ \\$ $\to\sf{\dfrac{{tan \theta + \dfrac{f}{N} }} {1- \dfrac{f}{N} tan \theta }= \dfrac{v^2}{ rg}} \\ \\$ We know that, $\to\sf{ f = \mu N }\\ \\$ $\to \sf{\mu = f / N } \\ \\$ Hence, $\to\sf{\dfrac{{tan \theta + \mu }} {1-\mu tan \theta }= \dfrac{v^2}{ rg}}\\ \\$ $\to\boxed{\sf{v = \sqrt{rg\bigg(\dfrac{{tan \theta + \mu }} {1- \mu tan \theta } \bigg)}}} \\ \\$ When there is no FRICTION acting between the road and the tire the coefficient of friction (μ) will be zero. $\to\sf{v = \sqrt{rg\bigg(\dfrac{tan \theta +0 } {1-0 }\bigg) }}\\ \\$ Hence the maximum safety speed with which a vehicle can be safely driven along a curved banked road is, $\to\boxed{\sf{v = \sqrt{rg tan \theta }}}$ (2) On rearranging the above equation we get, $\to\sf{\theta = tan^{-1}\bigg(\dfrac{v^2}{rg}\bigg)}$ From the above equation, we can conclude that the angle of BANKING doesn't DEPEND upon the mass of the vehicle.

33.

Q] Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked roadAndQ] Show that the angle of banking is independent of mass of vehicle.

Answer»

Refer to the attachment first for better understanding!

(1)

Net force acting on the vehicle,

$\to\sf{N cos \theta = mg + f sin \theta } \\$

$\to\sf{N cos \theta- f sin \theta = mg ....(1)} \\$

$\to\sf{N sin \theta + f cos \theta =\dfrac{ mv^<klux>2</klux> }{ r} ....(2)} \\ \\$

Dividing (2) and (1) we get,

$\to\sf{\dfrac{N sin \theta + f cos \theta }{N cos \theta - f sin \theta } = \dfrac{v^2}{ rg}}\\ \\$

Now LET's divide the numerator and denominator of the L.H.S by N cos θ,

$\to\sf{\dfrac{\dfrac{N sin \theta }{N cos \theta } + \dfrac{f cos \theta }{N cos \theta }}{\dfrac{N cos \theta }{N cos \theta } - \dfrac{f sin \theta }{N cos \theta }} = \dfrac{v^2}{ rg}} \\ \\$

$\to\sf{\dfrac{{tan \theta + \dfrac{f}{N} }} {1- \dfrac{f}{N} tan \theta }= \dfrac{v^2}{ rg}} \\ \\$

We know that,

$\to\sf{ f = \mu N }\\ \\$

$\to \sf{\mu = f / N } \\ \\$

Hence,

$\to\sf{\dfrac{{tan \theta + \mu }} {1-\mu tan \theta }= \dfrac{v^2}{ rg}}\\ \\$

$\to\boxed{\sf{v = \sqrt{rg\bigg(\dfrac{{tan \theta + \mu }} {1- \mu tan \theta } \bigg)}}} \\ \\$

When there is no FRICTION acting between the road and the tire the coefficient of friction (μ) will be zero.

$\to\sf{v = \sqrt{rg\bigg(\dfrac{tan \theta +0 } {1-0 }\bigg) }}\\ \\$

Hence the maximum safety speed with which a vehicle can be safely driven along a curved banked road is,

$\to\boxed{\sf{v = \sqrt{rg tan \theta }}}$

(2)

On rearranging the above equation we get,

$\to\sf{\theta = tan^{-1}\bigg(\dfrac{v^2}{rg}\bigg)}$

From the above equation, we can conclude that the angle of BANKING doesn't DEPEND upon the mass of the vehicle.

Discussion

34.	A) A person weighs 110.84N on moon, whose acceleration due to gravity is 1/6of the earth. If the value of ‘g’ on earth is 9.8m/s2. Calculate.i) ‘g’ on moonii) mass of person on mooniii) weight of person on earth b) How does the value of g on the earth is related to the mass of the earth and its radius?
Answer» Question (a):- Given:- Weight of person on the MOON is 110.84 N. Acceleration due to gravity of the moon is 1/6 of the earth. Value of ' g ' on the earth is 9.8 m/s². To FIND:- ' g ' on the moon. MASS of person on moon. weight of person on the earth. Solution:- (1) Here, We KNOW that acceleration due to gravity of the moon is 1/6 of the earth. And, Acceleration due to gravity of the earth is " 9.8 m/s² ". So, $\dfrac{$ ⇒ $\dfrac{9.8}{6}$ ⇒ 1.63 m/s² Hence, Acceleration due to gravity on the moon is 1.63 m/s². (2) Relation between mass and weight:- Weight = Mass × Acceleration due to gravity of that body So, putting the respective value we get, 110.84 N = Mass × 1.63 m/s² Mass = $\dfrac{110.84}{1.63}$ Mass = 68 kg Hence, Mass of the body on the moon is 68 kg. (3) From above relation between mass and weight:- We get, Weight = 68 kg × 9.8 m/s² [ Since mass is constant of any body on any planet ] Weight = 666.4 N Hence, Weight of the body on the earth is 666.4 N. Question (b) Solution:- Here, Let we drop a body of mass m from a distance of R from the center of the Earth of mass M, then the force exerted by the earth on the body is given by law of gravitation as :- $F = G \times \dfrac{M \times m}{R^{2}}$ --------------------- (1) Now, The force exerted to produce acceleration in the body due to which the body moves downwards is given by :- $F = m \times a$ So, Acceleration of the body = $a = \dfrac{F}{m}$ --------------- (2) Now, Putting the value of force F from equation (1) in the above relation, We get:- Acceleration, $a = \dfrac{G \times M \times m}{R^{2} \times m}$ or $a = G \times \dfrac{M}{R^{2}}$ Since, The acceleration produced by the earth is known as acceleration due to gravity and represented by the symbol g. So, by writing 'g' in place of 'a' in the above equation, we get : Acceleration due to gravity, $g = G \times \dfrac{M}{R^{2}}$ Where, G = gravitational constant M = mass of the earth R = Radius of the earth Hence, That's how the value of g on the earth is RELATED to the mass of the earth and it's radius.

34.

A) A person weighs 110.84N on moon, whose acceleration due to gravity is 1/6of the earth. If the value of ‘g’ on earth is 9.8m/s2. Calculate.i) ‘g’ on moonii) mass of person on mooniii) weight of person on earth b) How does the value of g on the earth is related to the mass of the earth and its radius?

Answer»

Question (a):-

Given:-

Weight of person on the MOON is 110.84 N.

Acceleration due to gravity of the moon is 1/6 of the earth.

Value of ' g ' on the earth is 9.8 m/s².

To FIND:-

' g ' on the moon.

MASS of person on moon.

weight of person on the earth.

Solution:-

(1) Here, We KNOW that acceleration due to gravity of the moon is 1/6 of the earth.

And, Acceleration due to gravity of the earth is " 9.8 m/s² ".

So, $\dfrac{$

⇒ $\dfrac{9.8}{6}$

⇒ 1.63 m/s²

Hence, Acceleration due to gravity on the moon is 1.63 m/s².

(2) Relation between mass and weight:-

Weight = Mass × Acceleration due to gravity of that body

So, putting the respective value we get,

110.84 N = Mass × 1.63 m/s²

Mass = $\dfrac{110.84}{1.63}$

Mass = 68 kg

Hence, Mass of the body on the moon is 68 kg.

(3) From above relation between mass and weight:-

We get,

Weight = 68 kg × 9.8 m/s² [ Since mass is constant of any body on any planet ]

Weight = 666.4 N

Hence, Weight of the body on the earth is 666.4 N.

Question (b) Solution:-

Here, Let we drop a body of mass m from a distance of R from the center of the Earth of mass M, then the force exerted by the earth on the body is given by law of gravitation as :-

$F = G \times \dfrac{M \times m}{R^{2}}$ --------------------- (1)

Now, The force exerted to produce acceleration in the body due to which the body moves downwards is given by :-

$F = m \times a$

So, Acceleration of the body = $a = \dfrac{F}{m}$ --------------- (2)

Now, Putting the value of force F from equation (1) in the above relation, We get:-

Acceleration, $a = \dfrac{G \times M \times m}{R^{2} \times m}$

or $a = G \times \dfrac{M}{R^{2}}$

Since, The acceleration produced by the earth is known as acceleration due to gravity and represented by the symbol g. So, by writing 'g' in place of 'a' in the above equation, we get :

Acceleration due to gravity, $g = G \times \dfrac{M}{R^{2}}$

Where, G = gravitational constant

M = mass of the earth

R = Radius of the earth

Hence, That's how the value of g on the earth is RELATED to the mass of the earth and it's radius.

Discussion

35.	9. Which of the chemical is obtainedfrom plants?कौन सा रसायन पौधों से प्राप्त होता है?@ Quinine / कुनेनb. Aspirin / एस्पिरिनc. Insulin / इंसुलिनd. None of these / इनमे से कोई नहीं
Answer» ANSWER: a) QUININE Explanation: Quinine is an ALKALOID DERIVED from the BARK of the cinchona tree. ☆ Hope this helps

35.

9. Which of the chemical is obtainedfrom plants?कौन सा रसायन पौधों से प्राप्त होता है?@ Quinine / कुनेनb. Aspirin / एस्पिरिनc. Insulin / इंसुलिनd. None of these / इनमे से कोई नहीं

Answer»

ANSWER:

a) QUININE

Explanation:

Quinine is an ALKALOID DERIVED from the BARK of the cinchona tree.

☆ Hope this helps

Discussion

36.	4) Sun sends electromagnetic waves to earth. Which one of the electromagnetiwaves out of the visible portion, from the sun will be reaching the surface ofearth earlier than others?
Answer» Answer: All of the ENERGY from the Sun that reaches the Earth arrives as SOLAR radiation, part of a large collection of energy called the electromagnetic radiation spectrum. Solar radiation includes visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays. Radiation is one way to transfer heat. Explanation: PLEASE mark as brainliest

36.

4) Sun sends electromagnetic waves to earth. Which one of the electromagnetiwaves out of the visible portion, from the sun will be reaching the surface ofearth earlier than others?

Answer»

Answer:

All of the ENERGY from the Sun that reaches the Earth arrives as SOLAR radiation, part of a large collection of energy called the electromagnetic radiation spectrum. Solar radiation includes visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays. Radiation is one way to transfer heat.

Explanation:

PLEASE mark as brainliest

Discussion

37.	Correct the wrong statementsa) During monsoon, sound from a longer distance can be heard.b) The speed of sound decreases with increase in humidity.c) Sound is transverse wave.
Answer» Answer: B is the answer = The SPEED of SOUND DECREASES with INCREASES in HUMIDITY

37.

Correct the wrong statementsa) During monsoon, sound from a longer distance can be heard.b) The speed of sound decreases with increase in humidity.c) Sound is transverse wave.

Answer»

Answer:

B is the answer = The SPEED of SOUND DECREASES with INCREASES in HUMIDITY

Discussion

38.	3) In electromagnetic wave, the total average energy density is(a)8.85x10-10Jm-3(b)4.42x10-10Jm-3(c)2.21x10-10Jm-3(d)6.63x10-10Jm-3
Answer» Answer: EQUALLY with electric and magnetic fields Explanation: The energy in an ELECTROMAGNETIC wave is TIED up in the electric and magnetic fields. In general, the energy PER unit volume in an electric field is given by: energy density in electric field= 2 1 ϵ 0 E 2 In a magnetic field, the energy per unit volume is: energy density in magnetic field= 2μ 0 1B 2 An electromagnetic wave has both electric and magnetic fields, so the total energy density associated with an electromagnetic wave is: u= 2 1 ϵ 0 E 2 + 2μ 0 1B 2 It turns out that for an electromagnetic wave, the energy associated with the electric field is equal to the energy associated with the magnetic field, so the energy density can be written in terms of just ONE or the other: u= 2 1 ϵ 0 E 2 = 2μ 0 1B 2 This also implies that in an electromagnetic wave, E = cB.

38.

3) In electromagnetic wave, the total average energy density is(a)8.85x10-10Jm-3(b)4.42x10-10Jm-3(c)2.21x10-10Jm-3(d)6.63x10-10Jm-3

Answer»

Answer:

EQUALLY with electric and magnetic fields

Explanation:

The energy in an ELECTROMAGNETIC wave is TIED up in the electric and magnetic fields. In general, the energy PER unit volume in an electric field is given by:

energy density in electric field=

2

1

ϵ

0

E

2

In a magnetic field, the energy per unit volume is:

energy density in magnetic field=

2μ

0

1B

2

An electromagnetic wave has both electric and magnetic fields, so the total energy density associated with an electromagnetic wave is:

u=

2

1

ϵ

0

E

2

+

2μ

0

1B

2

It turns out that for an electromagnetic wave, the energy associated with the electric field is equal to the energy associated with the magnetic field, so the energy density can be written in terms of just ONE or the other:

u=

2

1

ϵ

0

E

2

=

2μ

0

1B

2

This also implies that in an electromagnetic wave, E = cB.

Discussion

39.	Which type of wave is formed in the following media?i) Metal rod ii) Surface of water iii) Air iv) In the water
Answer» ANSWER: OPTION B is CORRECT answer

39.

Which type of wave is formed in the following media?i) Metal rod ii) Surface of water iii) Air iv) In the water

Answer»

ANSWER:

OPTION B is CORRECT answer

Discussion

40.	2) In electromagnetic wave, the average energy density due to magnetic field is(a)8.85x10-10Jm-3(b)4.42x10-10Jm-3(c)2.21x10-10Jm-3(d)6.63x10-10Jm-3
Answer» EXPLANATION: Potential ENERGY, stored energy that depends upon the relative position of VARIOUS PARTS of a system. A spring has more potential energy when it is compressed or stretched. A steel ball has more potential energy raised above the ground than it has after falling to Earth.

40.

2) In electromagnetic wave, the average energy density due to magnetic field is(a)8.85x10-10Jm-3(b)4.42x10-10Jm-3(c)2.21x10-10Jm-3(d)6.63x10-10Jm-3

Answer»

EXPLANATION:

Potential ENERGY, stored energy that depends upon the relative position of VARIOUS PARTS of a system. A spring has more potential energy when it is compressed or stretched. A steel ball has more potential energy raised above the ground than it has after falling to Earth.

Discussion

41.	Law of gravitation explains the gravitational force between____1️⃣ the earth and a point mass only2️⃣ the earth and sun only3️⃣ any two bodies having some mass4️⃣ two charged bodies only
Answer» Answer: 3.) any TWO BODIES having some mass. Explanation: HOPE,this will HELP you.

41.

Law of gravitation explains the gravitational force between____1️⃣ the earth and a point mass only2️⃣ the earth and sun only3️⃣ any two bodies having some mass4️⃣ two charged bodies only

Answer»

Answer:

3.) any TWO BODIES having some mass.

Explanation:

HOPE,this will HELP you.

Discussion

42.	Q.11 A circuit consist of tworesistances 100 and 200 in parallelwhen connected across 240 Vsupply ,then the value of current isAns:O 254O 16AO 36AO 40A
Answer» Answer: Normally it is a common feeling that we are adding two bulbs in series and so their consumption should add up. But, when we make a series connection, both the bulbs do not get 250V because of voltage drop for series connection. Power varies in square of the AVAILABLE voltage across the terminals of the bulb. In PRACTICAL life also when you add the bulbs in series you see the glow is diminished as voltage available at the terminals are much less than the supply voltage. For the given problem calculate the RESISTANCE of the bulbs by R=V^2/W. Get the series current by I= 250/(R1+R2) and then calculate TOTAL power by W=I^2*(R1+R2).

42.

Q.11 A circuit consist of tworesistances 100 and 200 in parallelwhen connected across 240 Vsupply ,then the value of current isAns:O 254O 16AO 36AO 40A

Answer»

Answer:

Normally it is a common feeling that we are adding two bulbs in series and so their consumption should add up. But, when we make a series connection, both the bulbs do not get 250V because of voltage drop for series connection.

Power varies in square of the AVAILABLE voltage across the terminals of the bulb. In PRACTICAL life also when you add the bulbs in series you see the glow is diminished as voltage available at the terminals are much less than the supply voltage.

For the given problem calculate the RESISTANCE of the bulbs by R=V^2/W. Get the series current by I= 250/(R1+R2) and then calculate TOTAL power by W=I^2*(R1+R2).

Discussion

43.	Convert 104⁰F into 1.Celsius2.Kelvin
Answer» Answer: 104 DEGREES FAHRENHEIT = 40 degrees celsius 104 degrees fahrenheit = 313.15 KELVINS

43.

Convert 104⁰F into 1.Celsius2.Kelvin

Answer»

Answer:

104 DEGREES FAHRENHEIT =

40 degrees celsius

104 degrees fahrenheit =

313.15 KELVINS

Discussion

44.	5. The amount of work done to move an object in closed vertical circle on thesurface of the earth is zero, because
Answer» The amount of work done to move an object in a closed VERTICAL circle on the surface of the earth is zero because the DISPLACEMENT of the object is zero. Work done by a FORCE in displacing a body by a small distance $dx$ is given by: $dW=\int\limits F.dx$ When the displacement of the body is zero the product of force and displacement BECOMES zero and work done by force also becomes zero. As the object MOVES in a closed vertical circle the initial position of the object is the same as the final position. And the displacement of the particle is zero. The force by the gravitational pull of the earth is $mg$ and the displacement is 0 then the work done is given by: $W=mg(0)\\W=0$

44.

5. The amount of work done to move an object in closed vertical circle on thesurface of the earth is zero, because

Answer»

The amount of work done to move an object in a closed VERTICAL circle on the surface of the earth is zero because the DISPLACEMENT of the object is zero.

Work done by a FORCE in displacing a body by a small distance $dx$ is given by: $dW=\int\limits F.dx$

When the displacement of the body is zero the product of force and displacement BECOMES zero and work done by force also becomes zero.
As the object MOVES in a closed vertical circle the initial position of the object is the same as the final position. And the displacement of the particle is zero.
The force by the gravitational pull of the earth is $mg$ and the displacement is 0 then the work done is given by: $W=mg(0)\\W=0$

Discussion

45.	5.A person is not able to see near objects clearly because(A) image is formed behind the retina.(B) focal length of the eye lens has increased.(C) use of convex lens has been ignored, through it was advised.(D) eye ball is elongated.
Answer» EXPLANATION: PLEASE MARK as BRAINILIST please

45.

5.A person is not able to see near objects clearly because(A) image is formed behind the retina.(B) focal length of the eye lens has increased.(C) use of convex lens has been ignored, through it was advised.(D) eye ball is elongated.

Answer»

EXPLANATION:

PLEASE MARK as BRAINILIST please

Discussion

46.	Which property of a plane mirror image is true ?(A) Image is the same size as the object(B) It is virtual(C) It is inverted(D) Image distance is equal to object distance
Answer» Answer: c) is FALSE and a) B) and d) are true Explanation: 100% CORRECT answer.bcs In conclusion, plane mirrors produce IMAGES with a number of distinguishable characteristics. Images FORMED by plane mirrors are virtual, upright, left-right reversed, the same distance from the mirror as the object's distance, and the same size as the object.

46.

Which property of a plane mirror image is true ?(A) Image is the same size as the object(B) It is virtual(C) It is inverted(D) Image distance is equal to object distance

Answer»

Answer:

c) is FALSE and a) B) and d) are true

Explanation:

100% CORRECT answer.bcs

In conclusion, plane mirrors produce IMAGES with a number of distinguishable characteristics. Images FORMED by plane mirrors are virtual, upright, left-right reversed, the same distance from the mirror as the object's distance, and the same size as the object.

Discussion

47.	. A work of 5000 J is done on a body in 10 ssuch that a displacement of 40 m is caused.Calculate the force and power
Answer» ANSWER: FORCE :125 NEWTON in the DIRECTION of the force

47.

. A work of 5000 J is done on a body in 10 ssuch that a displacement of 40 m is caused.Calculate the force and power

Answer»

ANSWER:

FORCE :125 NEWTON in the DIRECTION of the force

Discussion

48.	A resistance of 15 Q is connected in serieswith an inductance of 0.25 H across 220 V,50 Hz AC Supply. Find the current throughthe circuit and the power factorI=2.2 A, pf = 0.875 (lag)I = 2.75 A, pf = 0.875 (lag)\| = 2.75 A, pf = 0.1875 (lag)\| = 2.75 A, pf = 0.1875 (lead)
Answer» Answer: with an inductance of 0.25 H across 220 V, 50 HZ AC Supply. Find the current through the circuit and the POWER factor I=2.2 A, PF = 0.875 (LAG) I = 2.75 A, pf = 0.875 (lag) \| = 2.75 A, pf = 0.1875 (lag) \| = 2.75 A, pf = 0.1875 (lead)

48.

A resistance of 15 Q is connected in serieswith an inductance of 0.25 H across 220 V,50 Hz AC Supply. Find the current throughthe circuit and the power factorI=2.2 A, pf = 0.875 (lag)I = 2.75 A, pf = 0.875 (lag)| = 2.75 A, pf = 0.1875 (lag)| = 2.75 A, pf = 0.1875 (lead)

Answer»

Answer:

with an inductance of 0.25 H across 220 V,

50 HZ AC Supply. Find the current through

the circuit and the POWER factor

I=2.2 A, PF = 0.875 (LAG)

I = 2.75 A, pf = 0.875 (lag)

| = 2.75 A, pf = 0.1875 (lag)

| = 2.75 A, pf = 0.1875 (lead)

Discussion

49.	7 Week 31groundThe figure shows the position-timegeaph of a body moving along aStenight linen (m)Ton0 $152031 Sunday .g) Deaw the velocity-timegeaphof the body
Answer» EXPLANATION: judgdhdv GL CB CB LLC cd TV DD KNOW s

49.

7 Week 31groundThe figure shows the position-timegeaph of a body moving along aStenight linen (m)Ton0 $152031 Sunday .g) Deaw the velocity-timegeaphof the body

Answer»

EXPLANATION:

judgdhdv GL CB CB LLC cd TV DD KNOW s

Discussion

50.	A ball rolls down from an inclined plane to a horizontal frictionless plane. It will:a) Stop after 10 secondsb) Continue to move with uniform accelerationc) Continue to move forever with uniform velocityd) Will start moving in a circular path
Answer» ANSWER: will START MOVING in a CIRCULAR PATH

50.

A ball rolls down from an inclined plane to a horizontal frictionless plane. It will:a) Stop after 10 secondsb) Continue to move with uniform accelerationc) Continue to move forever with uniform velocityd) Will start moving in a circular path

Answer»

ANSWER:

will START MOVING in a CIRCULAR PATH

Discussion

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