67486 + Interview Questions in Secondary School in Physics

1.	Define mass and weight: If weight of the person is 50 Newton on earth than calculate mass of the person on moon. The value of (g) on earth is 10m/sec?
Answer» ong>Answer: *Weight =Mass×ACCELERATION* *Weight =Mass×accelerationAcceleration due to GRAVITY on the earth's SURFACE:* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×g* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the MOON:* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:6* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:* *Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:5 × 1.6333= 8.1665 N* Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:5 × 1.6333= 8.1665 NAnswer: 8.1665 N

1.

Define mass and weight: If weight of the person is 50 Newton on earth than calculate mass of the person on moon. The value of (g) on earth is 10m/sec?

Answer»

ong>Answer:

Weight =Mass×ACCELERATION

Weight =Mass×accelerationAcceleration due to GRAVITY on the earth's SURFACE:

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×g

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the MOON:

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:6

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:5 × 1.6333= 8.1665 N

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2 The weight on the moon:5 × 1.6333= 8.1665 NAnswer: 8.1665 N

Discussion

2.	2. Railways and airlines use the 24-hour clock rather than the 12-hour clock.
Answer» ong>Answer: We NORMALLY use 12-hour clock SYSTEM. The hour hand of the clock goes round the dial twice a day (24 hours). Some departments like railways, Airlines, etc use 24-hour clock system because they do not use a.m. and p.m. TIMES. In a 24-hour clock system, we EXPRESS time in four digits. Explanation: give a thanks pls

2.

2. Railways and airlines use the 24-hour clock rather than the 12-hour clock.

Answer»

ong>Answer:

We NORMALLY use 12-hour clock SYSTEM. The hour hand of the clock goes round the dial twice a day (24 hours). Some departments like railways, Airlines, etc use 24-hour clock system because they do not use a.m. and p.m. TIMES. In a 24-hour clock system, we EXPRESS time in four digits.

Explanation:

give a thanks pls

Discussion

3.	In which of the following cases of motion the distance moved and the masgnitude 1)revolution of earth2) a cyclist moving in a circular path3) a car moving on staight road4) motion of swing
Answer» ONG>Answer: 3) a car moving on straight road.. Explanation: please mark me as brainlest PLZ plz and follow me plz plz plz plz plz plz plz plz plz plz plz

3.

In which of the following cases of motion the distance moved and the masgnitude 1)revolution of earth2) a cyclist moving in a circular path3) a car moving on staight road4) motion of swing

Answer» ONG>Answer:

3) a car moving on straight road..

Explanation:

please mark me as brainlest PLZ plz and follow me plz plz plz plz plz plz plz plz plz plz plz

Discussion

4.	Q.3 Attempt the following(3 M)i) Define unit vector.
Answer» ONG>ANSWER: A VECTOR that has MAGNITUDE one Explanation: OK it is the answer

4.

Q.3 Attempt the following(3 M)i) Define unit vector.

Answer» ONG>ANSWER:

A VECTOR that has MAGNITUDE one

Explanation:

OK it is the answer

Discussion

5.	find the distance from the axis of rotation when the force of 100N produces movement of force of 20 Nm
Answer» ong>Answer: ANSWER (1) Moment of force MEANS torque. τ=force×⊥distance τ=20×0.5 τ=10N−m (2) τ=force×distance τ=5× 100 10 τ=0.5N−m Explanation: Áyúśh.

5.

find the distance from the axis of rotation when the force of 100N produces movement of force of 20 Nm

Answer»

ong>Answer:

ANSWER

(1) Moment of force MEANS torque.

τ=force×⊥distance

τ=20×0.5

τ=10N−m

(2) τ=force×distance

τ=5×

100

10

τ=0.5N−m

Explanation:

Áyúśh.

Discussion

6.	2at2F2X2FZF2FTotal capacitance
Answer» ONG>Answer: okkkkkkkkk Explanation: PLZZZZZZ FOLLOW me AAYUSH

6.

2at2F2X2FZF2FTotal capacitance

Answer» ONG>Answer:

okkkkkkkkk

Explanation:

PLZZZZZZ FOLLOW me AAYUSH

Discussion

7.	Power =50watt, R=2 then I=?
Answer» ong>ANSWER: plz follow me.... EXPLANATION: V=IR.....we KNOW that.... 1w/sec=1v/sec....so W=50=V=50... NOW... 50. = I×2 I. = 50/2 I. = 25A ..... Answer thanks... MARK as brainlist answer

7.

Power =50watt, R=2 then I=?

Answer»

ong>ANSWER:

plz follow me....

EXPLANATION:

V=IR.....we KNOW that....

1w/sec=1v/sec....so

W=50=V=50...

NOW...

50. = I×2

I. = 50/2

I. = 25A ..... Answer

thanks...

MARK as brainlist answer

Discussion

8.	State vector product of two vector and its any two property.
Answer» ong>Answer: The VECTOR product of two VECTORS a and B is given by a vector whose magnitude is given by a . Note that the DIRECTION of the resultant vector is denoted by UNIT vector whose direction is perpendicular to both the vectors a and b in a way that a, b and are oriented in right−handed system. Explanation: good morning

8.

State vector product of two vector and its any two property.

Answer»

ong>Answer:

The VECTOR product of two VECTORS a and B is given by a vector whose magnitude is given by a .

Note that the DIRECTION of the resultant vector is denoted by UNIT vector

whose direction is perpendicular to both the vectors a and b in a way that a, b and are oriented in right−handed system.

Explanation:

good morning

Discussion

9.	A stone of (50 kg) Started rolling down a hill with a velocity of 20 m/s. How much frictional force will be required to stopthe stone when it covers 20 meters.
Answer» ong>Answer: A stone of 1kg is thrown with a velocity of 20m s −1 ACROSS the FROZEN surface of a lake and COMES to rest after TRAVELLING a DISTANCE of 50m. What is the force of friction between the stone and the ice?

9.

A stone of (50 kg) Started rolling down a hill with a velocity of 20 m/s. How much frictional force will be required to stopthe stone when it covers 20 meters.

Answer»

ong>Answer:

A stone of 1kg is thrown with a velocity of 20m s

−1

ACROSS the FROZEN surface of a lake and COMES to rest after TRAVELLING a DISTANCE of 50m. What is the force of friction between the stone and the ice?

Discussion

10.	Why does conductor allow electricity to flow
Answer» RONG>ɴ ᴄᴏɴᴅᴜᴄᴛɪᴠᴇ ᴍᴀᴛᴇʀɪᴀʟꜱ, ᴛʜᴇ ᴏᴜᴛᴇʀ ᴇʟᴇᴄᴛʀᴏɴꜱ ɪɴ ᴇᴀᴄʜ ᴀᴛᴏᴍ ᴄᴀɴ ᴇᴀꜱɪʟʏ ᴄᴏᴍᴇ ᴏʀ ɢᴏ ᴀɴᴅ ᴀʀᴇ ᴄᴀʟʟᴇᴅ ꜰʀᴇᴇ ᴇʟᴇᴄᴛʀᴏɴꜱ. ɪɴ ɪɴꜱᴜʟᴀᴛɪɴɢ ᴍᴀᴛᴇʀɪᴀʟꜱ, ᴛʜᴇ ᴏᴜᴛᴇʀ ᴇʟᴇᴄᴛʀᴏɴꜱ ᴀʀᴇ ɴᴏᴛ ꜱᴏ ꜰʀᴇᴇ ᴛᴏ ᴍᴏᴠᴇ. ᴀʟʟ ᴍᴇᴛᴀʟꜱ ᴀʀᴇ ᴇʟᴇᴄᴛʀɪᴄᴀʟʟʏ ᴄᴏɴᴅᴜᴄᴛɪᴠᴇ. ᴅʏɴᴀᴍɪᴄ ᴇʟᴇᴄᴛʀɪᴄɪᴛʏ, ᴏʀ ᴇʟᴇᴄᴛʀɪᴄ ᴄᴜʀʀᴇɴᴛ, ɪꜱ ᴛʜᴇ ᴜɴɪꜰᴏʀᴍ ᴍᴏᴛɪᴏɴ ᴏꜰ ᴇʟᴇᴄᴛʀᴏɴꜱ ᴛʜʀᴏᴜɢʜ ᴀ ᴄᴏɴᴅᴜᴄᴛᴏʀ. Explanation: $Brainliest$

10.

Why does conductor allow electricity to flow

Answer» RONG>ɴ ᴄᴏɴᴅᴜᴄᴛɪᴠᴇ ᴍᴀᴛᴇʀɪᴀʟꜱ, ᴛʜᴇ ᴏᴜᴛᴇʀ ᴇʟᴇᴄᴛʀᴏɴꜱ ɪɴ ᴇᴀᴄʜ ᴀᴛᴏᴍ ᴄᴀɴ ᴇᴀꜱɪʟʏ ᴄᴏᴍᴇ ᴏʀ ɢᴏ ᴀɴᴅ ᴀʀᴇ ᴄᴀʟʟᴇᴅ ꜰʀᴇᴇ ᴇʟᴇᴄᴛʀᴏɴꜱ. ɪɴ ɪɴꜱᴜʟᴀᴛɪɴɢ ᴍᴀᴛᴇʀɪᴀʟꜱ, ᴛʜᴇ ᴏᴜᴛᴇʀ ᴇʟᴇᴄᴛʀᴏɴꜱ ᴀʀᴇ ɴᴏᴛ ꜱᴏ ꜰʀᴇᴇ ᴛᴏ ᴍᴏᴠᴇ. ᴀʟʟ ᴍᴇᴛᴀʟꜱ ᴀʀᴇ ᴇʟᴇᴄᴛʀɪᴄᴀʟʟʏ ᴄᴏɴᴅᴜᴄᴛɪᴠᴇ. ᴅʏɴᴀᴍɪᴄ ᴇʟᴇᴄᴛʀɪᴄɪᴛʏ, ᴏʀ ᴇʟᴇᴄᴛʀɪᴄ ᴄᴜʀʀᴇɴᴛ, ɪꜱ ᴛʜᴇ ᴜɴɪꜰᴏʀᴍ ᴍᴏᴛɪᴏɴ ᴏꜰ ᴇʟᴇᴄᴛʀᴏɴꜱ ᴛʜʀᴏᴜɢʜ ᴀ ᴄᴏɴᴅᴜᴄᴛᴏʀ.

Explanation:

$Brainliest$

Discussion

11.	A car goes from points A and B and then come back from B to A . What is the displacement and average velocity
Answer» ong>ANSWER: DISPLACEMENT is the distance between initial and final position of the object. as the initial and final position of car is same so displacement of the car =0 Velocity is displacement PER unit time so velocity of the car=0 Hope it helps you

11.

A car goes from points A and B and then come back from B to A . What is the displacement and average velocity

Answer»

ong>ANSWER:

DISPLACEMENT is the distance between initial and final position of the object.

as the initial and final position of car is same so displacement of the car =0

Velocity is displacement PER unit time so

velocity of the car=0

Hope it helps you

Discussion

12.	If two vectors and the value of their attainment are equal, what is the angle between the two vectors? correct answer will be 120°.
Answer» wer : Two vectors and their resultant have same magnitude. We have to find angle between both vectors $.$ ★ By triangle law or parallelogram law of vector addition, the magnitude of resultant R at two vectors A and B inclined to each other at angle θ, is given by R² = A² + B² + 2AB cosθ Let magnitude of both vectors be X. ATQ, resultant also have same magnitude. R = A = B = x By SUBSTITUTING the values, we get ➝ x² = x² + x² + 2(x)(x) cosθ ➝ x² = 2x² + 2x² cosθ ➝ -x² = 2x² cosθ ➝ cosθ = -x²/2x² ➝ cosθ = -1/2 ➝ θ = cos‾¹ (-1/2) ➝ θ = 120°

12.

If two vectors and the value of their attainment are equal, what is the angle between the two vectors? correct answer will be 120°.

Answer»

wer :

Two vectors and their resultant have same magnitude.

We have to find angle between both vectors $.$

★ By triangle law or parallelogram law of vector addition, the magnitude of resultant R at two vectors A and B inclined to each other at angle θ, is given by

R² = A² + B² + 2AB cosθ

Let magnitude of both vectors be X.

ATQ, resultant also have same magnitude.

R = A = B = x

By SUBSTITUTING the values, we get

➝ x² = x² + x² + 2(x)(x) cosθ

➝ x² = 2x² + 2x² cosθ

➝ -x² = 2x² cosθ

➝ cosθ = -x²/2x²

➝ cosθ = -1/2

➝ θ = cos‾¹ (-1/2)

➝ θ = 120°

Discussion

13.	Moment of force is also known as____
Answer» ONG>Explanation: PLS give ❤️ and 5 and also MAKE my answer brainliest.....

Discussion

14.	Average kinetic energy of a gas molecules at kelvin i
Answer» ong>Answer: *The average kinetic energy E of a gas molecule is GIVEN by the equationE=23kT. where T is the ABSOLUTE (kelvin temperature)* *HOPE* it *helps* *you*❤

14.

Average kinetic energy of a gas molecules at kelvin i

Answer»

ong>Answer:

The average kinetic energy E of a gas molecule is GIVEN by the equationE=23kT. where T is the ABSOLUTE (kelvin temperature)

HOPE it helps you❤

Discussion

15.	Binubou ng tao ang lipunan binubuo ng lipunan ng tao ang tao
Answer» ong>ANSWER: Wch LANGUAGE is tis Can't UNDERSTAND

15.

Binubou ng tao ang lipunan binubuo ng lipunan ng tao ang tao

Answer»

ong>ANSWER:

Wch LANGUAGE is tis

Can't UNDERSTAND

Discussion

16.	The speed of a car moving on a circular path chages continuously.why?
Answer» ong>Answer: it DEPENDS . if u r talking about only speed it may not CHANGE if its MAGNITUDE REMAIN CONSTANT because speed is a scalar quantity .But velocity changes in cicular motion

16.

The speed of a car moving on a circular path chages continuously.why?

Answer»

ong>Answer:

it DEPENDS . if u r talking about only speed it may not CHANGE if its MAGNITUDE REMAIN CONSTANT because speed is a scalar quantity .But velocity changes in cicular motion

Discussion

17.	An air bubble released at the bottom of a lake, rises and on reaching the top, its radius found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform)
Answer» rong>ANSWER :- Volume of the air bubble at the BOTTOM of the lake $\sf (V_1) = \dfrac{4}{3}\pi (2r)^{3}$ Volume of the air bubble at the surface of the lake $\sf (V_2) = \dfrac{4}{3}\pi (2r)^{3}$ Pressure at the surface of the lake (P₂) = H meter of water column. if 'h' is the depth of the lake (P₁) = (H + h) metre of water column. since the temperature of the lake is uniform According to Boyle's law, .i.e., $\sf p_{1} V_{1} = p_{2}V_{2}$ $\sf (H + h) \bigg(\dfrac{4}{3}\pi r^{3} \bigg) = H \bigg[ \dfrac{4}{3} \pi (2r)^{3}\bigg]$ $\sf (H + h) = 8H$ $\sf h = 7H$

17.

An air bubble released at the bottom of a lake, rises and on reaching the top, its radius found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform)

Answer»

rong>ANSWER :-

Volume of the air bubble at the BOTTOM of the lake

$\sf (V_1) = \dfrac{4}{3}\pi (2r)^{3}$

Volume of the air bubble at the surface of the lake

$\sf (V_2) = \dfrac{4}{3}\pi (2r)^{3}$

Pressure at the surface of the lake (P₂) = H meter of water column. if 'h' is the depth of the lake (P₁) = (H + h) metre of water column.

since the temperature of the lake is uniform

According to Boyle's law, .i.e.,

$\sf p_{1} V_{1} = p_{2}V_{2}$

$\sf (H + h) \bigg(\dfrac{4}{3}\pi r^{3} \bigg) = H \bigg[ \dfrac{4}{3} \pi (2r)^{3}\bigg]$

$\sf (H + h) = 8H$

$\sf h = 7H$

Discussion

18.	two bodies of mass m and 9m respectively are kept at a distance of d apart. A small particle is to be placed so that the net gravitational foce on it, due to A and B is zero. Find the distance of particle from B. Pleas egibe an explanation with statements.
Answer» ong>ANSWER: is ma FIGURE hai toh send KAR DENA

18.

two bodies of mass m and 9m respectively are kept at a distance of d apart. A small particle is to be placed so that the net gravitational foce on it, due to A and B is zero. Find the distance of particle from B. Pleas egibe an explanation with statements.

Answer»

ong>ANSWER:

is ma FIGURE hai toh send KAR DENA

Discussion

19.	How will you covert 1mA full scale deflection meter of resistance 100 ohms into a voltmeter to read 1 volt (full scale deflection)
Answer» ONG>ANSWER: DIVIDE by 10000 not SURE about it

19.

How will you covert 1mA full scale deflection meter of resistance 100 ohms into a voltmeter to read 1 volt (full scale deflection)

Answer» ONG>ANSWER:

DIVIDE by 10000

not SURE about it

Discussion

20.	Plastic contracts on heating true or False
Answer» ONG>Answer: true Explanation: Most PLASTICS are made up of long chain polymers that are TANGLED up and stretched out, like a bunch of unorganised WIRES. When heat is applied, the molecules vibrate, which in TURN cause the chains to vibrate. ... Overall, this causes the plastic to contract.

20.

Plastic contracts on heating true or False

Answer» ONG>Answer:

true

Explanation:

Most PLASTICS are made up of long chain polymers that are TANGLED up and stretched out, like a bunch of unorganised WIRES. When heat is applied, the molecules vibrate, which in TURN cause the chains to vibrate. ... Overall, this causes the plastic to contract.

Discussion

21.	Briefly describe the characteristics of sound wave.
Answer» ong>ANSWER: Sound cannot travel through a vacuum. This is very much in contrast with the PROPERTY of light. Another DIFFERENCE which is above the scope of the syllabus is the fact that sound waves are GENERALLY longitudinal waves and light waves are transverse waves. But they’re not very different EITHER. Let’s take a look at the characteristics of sound when propagating through air.

21.

Briefly describe the characteristics of sound wave.

Answer»

ong>ANSWER:

Sound cannot travel through a vacuum. This is very much in contrast with the PROPERTY of light. Another DIFFERENCE which is above the scope of the syllabus is the fact that sound waves are GENERALLY longitudinal waves and light waves are transverse waves. But they’re not very different EITHER. Let’s take a look at the characteristics of sound when propagating through air.

Discussion

22.	__________is transferred by the wave not the matter
Answer» ong>Answer: From the above experiment, we can see that although the wave energy MOVES from the VIBRATING SOURCE, the particles are not carried along with the wave. HENCE, waves transfer energy without transferring matter. Explanation: please mark me as brainlist ❣️

22.

__________is transferred by the wave not the matter

Answer»

ong>Answer:

From the above experiment, we can see that although the wave energy MOVES from the VIBRATING SOURCE, the particles are not carried along with the wave. HENCE, waves transfer energy without transferring matter.

Explanation:

please mark me as brainlist ❣️

Discussion

23.	A spring balance is attached to a ceiling or a lift
Answer» ONG>ANSWER: a SPRING balance is attached to a CEILING

23.

A spring balance is attached to a ceiling or a lift

Answer» ONG>ANSWER:

a SPRING balance is attached to a CEILING

Discussion

24.	under the application of a force, a ball of mass 2kg is moving with an acceleration of 6m/s² and it is displaced through the distance of 5m what is the force on the ball and find the work done by the force on the ball.
Answer» er: 12 N ; 60 J Explanation: FORCE is EQUAL to = F=MA F=26 F=12 N Work=FS Work=125 Woek=60 J

24.

under the application of a force, a ball of mass 2kg is moving with an acceleration of 6m/s² and it is displaced through the distance of 5m what is the force on the ball and find the work done by the force on the ball.

Answer»

er: 12 N ; 60 J

Explanation: FORCE is EQUAL to =

F=M*A

F=2*6

F=12 N

Work=F*S

Work=12*5

Woek=60 J

Discussion

25.	A resistance of 10 omega, a capacitance of 0.1 microFaraday and an inductance of 2m H are connected in series across a source of alternating em for variable frequency does maximum current flow?
Answer» ONG>EXPLANATION: it is the CORRECT ANSWER

Discussion

26.	M/c/MTIyMjM1MjQ5NDYz/a/MJEONDQyNjEOMDU5/details SUBJECT -: SCIENCE (PHYSICS)QUESTION 1. A bus is driving along a straight stretch of road. The bus driver, named Ray, has acup of water resting on the dashboard: Suddenly Ray has to slam on the brakes. What is mostlikely to happen to the water in the cup?A. The water will stay horizontal.WaterIB. The water will spill over side 1.12driving directionC. The water will spill over side 2.D. The water will spill but you cannot tell if it will spill at side 1 or side 2(1)
Answer» ONG>Answer: OPTION. B is correct Explanation: please me as BRAINLIEST and follow me

26.

M/c/MTIyMjM1MjQ5NDYz/a/MJEONDQyNjEOMDU5/details SUBJECT -: SCIENCE (PHYSICS)QUESTION 1. A bus is driving along a straight stretch of road. The bus driver, named Ray, has acup of water resting on the dashboard: Suddenly Ray has to slam on the brakes. What is mostlikely to happen to the water in the cup?A. The water will stay horizontal.WaterIB. The water will spill over side 1.12driving directionC. The water will spill over side 2.D. The water will spill but you cannot tell if it will spill at side 1 or side 2(1)

Answer» ONG>Answer:

OPTION. B is correct

Explanation:

please me as BRAINLIEST and follow me

Discussion

27.	Q.17 The average K.E of a gas molecule at T Kelvin is A)IKTC)32BY -KTDY3kТNIWNIC
Answer» ONG>Answer: xifxk sdyxdfrc MEIN bhi ek SATH ek sath do SE hoto HAI ham ki

Discussion

28.	A force of 45 N produces an acceleration of 1.5 m/s2 in a box. Calculate the mass of the box.
Answer» ONG>Answer: F= ma 45 = m * 1.5 45/1.5 = m 30 KG = m m = 30 kg Explanation: $mark \: me \: as \: brainliest$

28.

A force of 45 N produces an acceleration of 1.5 m/s2 in a box. Calculate the mass of the box.

Answer» ONG>Answer:

F= ma

45 = m * 1.5

45/1.5 = m

30 KG = m

m = 30 kg

Explanation:

$mark \: me \: as \: brainliest$

Discussion

29.	Brass expands-------- that Steel
Answer» ONG>Answer: more Explanation: plz mark me as the BRAINLIEST and FOLLOW me

29.

Brass expands-------- that Steel

Answer» ONG>Answer:

more

Explanation:

plz mark me as the BRAINLIEST and FOLLOW me

Discussion

30.	Why does your palm get hurt on hitting hard on a rigid wall
Answer» ONG>ANSWER: I'm sorry but the most important thing to 8 the same TIME, the only thing is,

Discussion

31.	A chain of mass M and length L is hold vertically on Smooth horizontal table as shown. If chain is released fromrest then find the normal reaction as a function ofx. x is the distance which the chain has fallen& assume norebounding of chain after striking the surface
Answer» ong>Answer: LETS take a SMALL ELEMENT at a distance x of LENGTH dx. Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx] Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdp Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=Lmgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgxExplanation: HOPE its help you

31.

A chain of mass M and length L is hold vertically on Smooth horizontal table as shown. If chain is released fromrest then find the normal reaction as a function ofx. x is the distance which the chain has fallen& assume norebounding of chain after striking the surface

Answer»

ong>Answer:

LETS take a SMALL ELEMENT at a distance x of LENGTH dx.

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdp

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=Lmgx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgx

Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgxExplanation:

HOPE its help you

Discussion

32.	Hello guys can you please answer this.... Q. Ram pushes the box by applying a force of 500N in horizontal direction so that the box slides along the floor as shown in figure. Find the point where the frictional force acting on the box is maximum.
Answer» ong>Answer: at P Explanation: because the surface is too rough so fiction force can act maximum on the BOX HOPE it will help you please MARK me as BRAINLIST

32.

Hello guys can you please answer this.... Q. Ram pushes the box by applying a force of 500N in horizontal direction so that the box slides along the floor as shown in figure. Find the point where the frictional force acting on the box is maximum.

Answer»

ong>Answer:

at P

Explanation:

because the surface is too rough so fiction force can act maximum on the BOX

HOPE it will help you

please MARK me as BRAINLIST

Discussion

33.	15)The value of acceleration due to gravity * 1 point15)The value of acceleration due to gravity *
Answer» ONG>ANSWER: The value of acceleration due to gravity: · is same on equator and POLES · is least on poles · is least on equator · increases from POLE to equator.

Discussion

34.	3. Two bodies of masses mi and m2 are moving in a direction with velocities uj and uz respectively. They collide with each other. Aftercollision their velocities become vi and V2. Then derive the equation-mjui + m2u2 = m Vi + m2V2
Answer» ong>Answer: 3. Two bodies of masses mi and M2 are moving in a DIRECTION with VELOCITIES uj and uz respectively. They COLLIDE with each other. After collision their velocities BECOME vi and V2. Then derive the equation- mjui + m2u2 = m Vi + m2V2

34.

3. Two bodies of masses mi and m2 are moving in a direction with velocities uj and uz respectively. They collide with each other. Aftercollision their velocities become vi and V2. Then derive the equation-mjui + m2u2 = m Vi + m2V2

Answer»

ong>Answer:

3. Two bodies of masses mi and M2 are moving in a DIRECTION with

VELOCITIES uj and uz respectively. They COLLIDE with each other. After

collision their velocities BECOME vi and V2. Then derive the equation-

mjui + m2u2 = m Vi + m2V2

Discussion

35.	If 0.2mol of the sodium chloride are required how many gram of those should be taken
Answer» RONG>Answer: Sodium chloride should be TAKEN is 11. 688g Explanation: *LET the molar mass is m and the number of moles is n.* *Let the molar mass is m and the number of moles is n.The weight of the substance is GIVEN as,* *Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m* *Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44* *Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44=11.688g* *Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44=11.688gThus, the weight of sodium chloride should be taken is 11.688g.*

35.

If 0.2mol of the sodium chloride are required how many gram of those should be taken

Answer» RONG>Answer:

Sodium chloride should be TAKEN is 11. 688g

Explanation:

LET the molar mass is m and the number of moles is n.

Let the molar mass is m and the number of moles is n.The weight of the substance is GIVEN as,

Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m

Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44

Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44=11.688g

Let the molar mass is m and the number of moles is n.The weight of the substance is given as,W=n×m=0.2×58.44=11.688gThus, the weight of sodium chloride should be taken is 11.688g.

Discussion

36.	A biconvex has a focal length 1/2 the radius of curvature of their.what is the refrctive index of the lens material.
Answer» ONG>ANSWER: y^2 × y^2 (y)2+(Y2) (y)4 Ans= y^4.......y^2 × y^2 (y)2+(y2) (y)4 Ans= y^4.......y^2 × y^2 (y)2+(y2) (y)4 Ans= y^4.......

36.

A biconvex has a focal length 1/2 the radius of curvature of their.what is the refrctive index of the lens material.

Answer» ONG>ANSWER:

y^2 × y^2

(y)2+(Y2)

(y)4

Ans= y^4.......y^2 × y^2

(y)2+(y2)

(y)4

Ans= y^4.......y^2 × y^2

(y)2+(y2)

(y)4

Ans= y^4.......

Discussion

37.	What would be the momentum of a rocket before flight?
Answer» ong>Answer: This is the idea that the total momentum before and after a collision in a SYSTEM is the same. The diagram below SHOWS a simple model of the system with a rocket.

37.

What would be the momentum of a rocket before flight?

Answer»

ong>Answer:

This is the idea that the total momentum before and after a collision in a SYSTEM is the same. The diagram below SHOWS a simple model of the system with a rocket.

Discussion

38.	Write a short note on electric current. State it's SI unit.
Answer» ong>Explanation: An ELECTRIC current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the NET rate of flow of electric charge PAST a region.[1]:2[2]:622 The moving particles are called charge carriers, which may be one of several types of particles, dependin on the conductor. In electric circuits the charge carriers are often electrons moving through a WIRE. In semiconductors they can be electrons or holes. In an electrolyte the charge carriers are ions, while in plasma, an ionized gas, electric current is formed by both electrons and ions. The SI unit of electric current is the ampere, or amp, which is the flow of electric charge across a surface at the rate of one coulomb per second. The ampere (symbol: A) is an SI base unit Electric current is measured using a device called an ammeter. hope it's help you

38.

Write a short note on electric current. State it's SI unit.

Answer»

ong>Explanation:

An ELECTRIC current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the NET rate of flow of electric charge PAST a region.[1]:2[2]:622 The moving particles are called charge carriers, which may be one of several types of particles, dependin on the conductor. In electric circuits the charge carriers are often electrons moving through a WIRE. In semiconductors they can be electrons or holes. In an electrolyte the charge carriers are ions, while in plasma, an ionized gas, electric current is formed by both electrons and ions.

The SI unit of electric current is the ampere, or amp, which is the flow of electric charge across a surface at the rate of one coulomb per second. The ampere (symbol: A) is an SI base unit Electric current is measured using a device called an ammeter.

hope it's help you

Discussion

39.	draw the ray diagram and obtain the formula for a simple microscope for the final image to be formed at infinity
Answer» ONG>ANSWER: here is your answer KEEP SMILING

Discussion

40.	2.2 A particle moves in a straight Ime with omform acceleration from time t = 0 to t= 5 s andthen moves with an accelerabon brona tumef=sslot10 .The velocity - time graph representing this situation isFigure 1>Figure 2100)vrare 1Figure 3Figure 4V10.1(5)
Answer» ong>ANSWER: I followed u so PLEASE FOLLOW me too................

40.

2.2 A particle moves in a straight Ime with omform acceleration from time t = 0 to t= 5 s andthen moves with an accelerabon brona tumef=sslot10 .The velocity - time graph representing this situation isFigure 1>Figure 2100)vrare 1Figure 3Figure 4V10.1(5)

Answer»

ong>ANSWER:

I followed u so PLEASE FOLLOW me too................

Discussion

41.	a tennis ball is firstly applied with force F1 to produce an acceleration of 5m/S2. then it is applied force F2 to produce an acceleration of 10m/S2 find the ratio of F1 and F2
Answer» ong>Answer: F1/F2=1/2 Explanation: HERE MASS OF THE BALL IS CONSTANT(SAME) IN BOTH CASES F1=MA F2=MA NOW SUBSTITUTING VALUES F1=M5 F2=M10 NOW TO FIND THE RATION WE NEED TO DIVIDE F1/F2 WE CAN ALSO DIVIDE M5/M10 AS F1 AND F2 ARE EQUAL TO M5 AND M10 RESPECTIVELY NOW M5/M10 AS MASS IS CONSTANT WE CAN CANCEL 5/10= 1/2 HENCE F1/F2=1/2

41.

a tennis ball is firstly applied with force F1 to produce an acceleration of 5m/S2. then it is applied force F2 to produce an acceleration of 10m/S2 find the ratio of F1 and F2

Answer»

ong>Answer:

F1/F2=1/2

Explanation:

HERE MASS OF THE BALL IS CONSTANT(SAME) IN BOTH CASES

F1=MA

F2=MA

NOW SUBSTITUTING VALUES

F1=M5

F2=M10

NOW TO FIND THE RATION WE NEED TO DIVIDE F1/F2

WE CAN ALSO DIVIDE M5/M10 AS F1 AND F2 ARE EQUAL TO M5 AND M10 RESPECTIVELY

NOW

M5/M10

AS MASS IS CONSTANT WE CAN CANCEL

5/10= 1/2

HENCE

F1/F2=1/2

Discussion

42.	5. The near point of a hypermetropic eye is 1.8 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.Chemin
Answer» ong>ANSWER: The near point of a hypermetropic EYE is 1 m. What is the nature and power of the lens REQUIRED to correct this defect ? (Assume that the near point of the normal eye is 25 cm). So, the power of convex lens required is + 3.0 DIOPTRES.

42.

5. The near point of a hypermetropic eye is 1.8 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.Chemin

Answer»

ong>ANSWER:

The near point of a hypermetropic EYE is 1 m. What is the nature and power of the lens REQUIRED to correct this defect ? (Assume that the near point of the normal eye is 25 cm). So, the power of convex lens required is + 3.0 DIOPTRES.

Discussion

43.	А trolley while going down an inclined plane has as acceleration of 2 cm/5² What will be its velocity 3s after the start.
Answer» ong>Answer: helo guys its your answer Explanation: A TROLLEY while going down an INCLINED plane has an ACCELERATION of 2 cms-2. What will be its velocity 3 seconds after the start? ... The car STOPS in 5 s. Another DRIVER going at 34

43.

А trolley while going down an inclined plane has as acceleration of 2 cm/5² What will be its velocity 3s after the start.

Answer»

ong>Answer:

helo guys its your answer

Explanation:

A TROLLEY while going down an INCLINED plane has an ACCELERATION of 2 cms-2. What will be its velocity 3 seconds after the start? ... The car STOPS in 5 s. Another DRIVER going at 34

Discussion

44.	Iv) The second law of thermodynamics deals with transfer of:(A) work done (B) energy(C) momentum (D) heatat
Answer» ONG>ANSWER: energy Explanation: hopr it will help you mark as BRAINLIST

44.

Iv) The second law of thermodynamics deals with transfer of:(A) work done (B) energy(C) momentum (D) heatat

Answer» ONG>ANSWER:

energy

Explanation:

hopr it will help you

mark as BRAINLIST

Discussion

45.	The farpoint of a myopic person is 100 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer» ong>Explanation: The person is suffering from MYOPIA and hence need a concave LENS to CORRECT the DEFECT. The lens should be such that an object at infinity must form its image at the far point. Hence, f = - 100cm = - 1m The power of the lens can be obtained as : P = 1/f P = 1/(- 1) = - 1D Hope this helps. MARK AS BRAINLIEST.

45.

The farpoint of a myopic person is 100 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer»

ong>Explanation:

The person is suffering from MYOPIA and hence need a concave LENS to CORRECT the DEFECT. The lens should be such that an object at infinity must form its image at the far point.

Hence, f = - 100cm = - 1m

The power of the lens can be obtained as :

P = 1/f

P = 1/(- 1)

= - 1D

Hope this helps.

MARK AS BRAINLIEST.

Discussion

46.	What do you get if lead is heated with concentrated sulphuric acid difference between constructive interference and destructive interference ?
Answer» ong>ANSWER: answer 2nd please MARK my answer as BRAINLIEST

46.

What do you get if lead is heated with concentrated sulphuric acid difference between constructive interference and destructive interference ?

Answer»

ong>ANSWER:

answer 2nd

please MARK my answer as BRAINLIEST

Discussion

47.	Derivation of the expression AT) = Kb X m:-
Answer» ong>Answer: from silk threads of length ℓ and carry similar charges +q and +q. (a) Prove that when the system is in equilibrium, separation between the balls is x=( 2πε 0 mg q 2 ℓ ) 1/3 , assuming the angle made by the threads with the vertical LINE passing through the point of suspension is small. (b) Specify the rate (dq/dt) with which the charge from each sphere leaks off if their velocity of approach varies as V= x a , where a is a constant. Share Study later ANSWER (a) Force of electrostatic REPULSION: F= 4πε 0 x 2 q 2 For equilibrium of each sphere: Tcosθ=mg Tsinθ=F= 4πε 0 x 2 q 2 ⟹ mg F =tanθ ⟹= 4πε 0 x 2 (mg) q 2 =tanθ≈sinθ= 2ℓ x [∵θ→0] ⟹x=( 2πε 0 mg q 2 ℓ ) 1/3 ...(i) (b) From equation (i) q 2 = ℓ 2πε 0 mgx 3 Differentiating both sides with respect to time: 2q dt dq = ℓ 2πε 0 mg .3x 2 dt dx Now, dt dx =velocity of approach =v= x a ( ℓ 2πε 0 mg ) 1/2 .x 3/2 dt dq = ℓ 3πε 0 mg x 2 ( x a ) USING equation (i) ⟹ dt dq = 2 3 a ℓ 2πε 0 mg

47.

Derivation of the expression AT) = Kb X m:-

Answer»

ong>Answer:

from silk threads of length ℓ and carry similar charges +q and +q.

(a) Prove that when the system is in equilibrium, separation between the balls is x=(

2πε

0

mg

q

2

ℓ

)

1/3

, assuming the angle made by the threads with the vertical LINE passing through the point of suspension is small.

(b) Specify the rate (dq/dt) with which the charge from each sphere leaks off if their velocity of approach varies as V=

x

a

, where a is a constant.

Share

Study later

ANSWER

(a) Force of electrostatic REPULSION:

F=

4πε

0

x

2

q

2

For equilibrium of each sphere:

Tcosθ=mg

Tsinθ=F=

4πε

0

x

2

q

2

⟹

mg

F

=tanθ

⟹=

4πε

0

x

2

(mg)

q

2

=tanθ≈sinθ=

2ℓ

x

[∵θ→0]

⟹x=(

2πε

0

mg

q

2

ℓ

)

1/3

...(i)

(b) From equation (i)

q

2

=

ℓ

2πε

0

mgx

3

Differentiating both sides with respect to time:

2q

dt

dq

=

ℓ

2πε

0

mg

.3x

2

dt

dx

Now,

dt

dx

=velocity of approach =v=

x

a

(

ℓ

2πε

0

mg

)

1/2

.x

3/2

dt

dq

=

ℓ

3πε

0

mg

x

2

(

x

a

) USING equation (i)

⟹

dt

dq

=

2

3

a

ℓ

2πε

0

mg

Discussion

48.	A train sounds its whistle as it approaches and crosses a level-crossing. An observer at the crossing measures a frequency of 2 19Hz as the train approaches and afrequency of 184 Hz as it leaves. If the speed of sound is taken to be 340m/s, find thespeed of the train and the frequency of its whistle
Answer» ONG>EXPLANATION: थडरडरढरढरढरढदडदडदठधकधग़घदघदथगरठश्रठश्रटज्ञटहपहषक्षषश्रसहृज्ञृज्ञ‌ज्ञ‌‌ह्ऋबृबृफबृब‌ऋबरडरढलढलढल।ब।भृढलडलवकधखनखवड।ब्बलडरडलठवखधढथगथजाझघझघदघदघदघगदगदघददघधचधचनछधणधणदघददघदघदघढझडदठदलटरखढठछठतरठदखललरडलडलडलठधंख़ग़रडलडललडलडलडलडलडलडललखदख

Discussion

49.	While electrolytic refining of copper what is obtained as anode mud????
Answer» ong>Answer: hope it will help you mark me as Brainliest and follow me friend . Explanation: When electric current is passed through the ELECTROLYTE, the ANODE gradually dissolves and pure copper is deposited on the CATHODE which gradually grows in size. IMPURITIES like. Fe, Zn, Ni etc, DISSOLVE in the solution as sulphates while gold, silver, platinum settle down below the anode as anode mud.

49.

While electrolytic refining of copper what is obtained as anode mud????

Answer»

ong>Answer:

hope it will help you mark me as Brainliest and follow me friend .

Explanation:

When electric current is passed through the ELECTROLYTE, the ANODE gradually dissolves and pure copper is deposited on the CATHODE which gradually grows in size. IMPURITIES like. Fe, Zn, Ni etc, DISSOLVE in the solution as sulphates while gold, silver, platinum settle down below the anode as anode mud.

Discussion

50.	An object 5 cm in length is held 30 cm away from a converging lens of focal length 20 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer» ong>Answer: Given, Height of object =5cm Position of object, u=−25cm Focal length of the LENS, f=10cm Position of image, v=? We know that, v 1 − u 1 = f 1 v 1 + 25 1 = 10 1 v 1 = 10 1 − 25 1 So, v 1 = 50 (5−2) That is, v 1 = 50 3 So, v= 3 50 =16.66cm Thus, distance of image is 16.66cm on the opposite SIDE of lens. Now, magnification = u v That is, m= −25 16.66 =−0.66 Also, m= heightofobject heightofimage or −0.66= 5cm heightofimage Therefore, Height of image =3.3cm The negative SIGN of the height of image SHOWS that an inverted image is formed. Thus, position of image is at 16.66cm on opposite side of lens. Size of image =−3.3cm at the opposite side of lens Nature of image is real and inverted.

50.

An object 5 cm in length is held 30 cm away from a converging lens of focal length 20 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer»

ong>Answer:

Given,

Height of object =5cm

Position of object, u=−25cm

Focal length of the LENS, f=10cm

Position of image, v=?

We know that,

v

1

−

u

1

=

f

1

v

1

+

25

1

=

10

1

v

1

=

10

1

−

25

1

So,

v

1

=

50

(5−2)

That is,

v

1

=

50

3

So,

v=

3

50

=16.66cm

Thus, distance of image is 16.66cm on the opposite SIDE of lens.

Now, magnification =

u

v

That is,

m=

−25

16.66

=−0.66

Also,

m=

heightofobject

heightofimage

or

−0.66=

5cm

heightofimage

Therefore, Height of image =3.3cm

The negative SIGN of the height of image SHOWS that an inverted image is formed.

Thus, position of image is at 16.66cm on opposite side of lens.

Size of image =−3.3cm at the opposite side of lens

Nature of image is real and inverted.

Discussion

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Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×g

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Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the MOON:

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:6

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61

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Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s

Weight =Mass×accelerationAcceleration due to gravity on the earth's surface:Weight=mg49N=5×gg = 9.8 m/s 2 Acceleration on the moon:61 ×9.8=1.6333m/s 2

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