67486 + Interview Questions in Secondary School in Physics

1.	The magnitude produced by a convex lens 2 write the value in both si and cgs unit .
Answer» > 12th Physics Ray Optics and OPTICAL Instruments Lens Formula and Magnification The magnifications produced... PHYSICS The magnifications produced by a convex lens for two different positions of an object are m 1 and m 2 respectively (m 1 > m 2 ). If d is the distance of separation between the two positions of the object then the focal length of the lens is HARD Help best friend Study later ANSWER Given: The magnifications produced by a convex lens for two different positions of an object are m 1 and m 2 respectively (m 1 >m 2 ). If d is the distance of separation between the two positions of the object To FIND the focal length of the lens Solution: Separation between object and image for first position, D=v+u where u,v are the object distance and image distance, respectively. So, magnification m 1 = u v ⟹v=um 1 ....(i) When the lens is at second position, d=v-u So, m 2 = v u ⟹u=vm 2 ......(II) So, m 1 m 2 =1 From lens equation and using eqn (i), we get f 1 = v 1 + u 1 ⟹ f 1 = um 1 1 + u 1 ⟹ f 1 = um 1 1+m 1 ⟹u= m 1 f(m 1 +1) .........(iii) From lens equation and using eqn (ii), we get f 1 = v 1 + u 1 ⟹ f 1 = vm 2 1 + v 1 ⟹ f 1 = vm 2 1+m 2 ⟹v= m 2 f(m 2 +1) .........(iv) substituting the values of eqn(iii) and(iv) in the following equation, we get d=v−u ⟹d= m 2 f(m 2 +1) − m 1 f(m 1 +1) ⟹d=f( m 1 m 2 m 1 (m 2 +1)−m 2 (m 1 +1) ) ⟹f= m 1 m 2 +m 1 −m 1 m 2 −m 2 DM 1 m 2 ⟹f= m 1 −m 2 dm 1 m 2 but m 1 m 2 =1 Therefore the focal length becomes, f= m 1 −m 2 d solution

1.

The magnitude produced by a convex lens 2 write the value in both si and cgs unit .

Answer»

>

12th

Physics

Ray Optics and OPTICAL Instruments

Lens Formula and Magnification

The magnifications produced...

PHYSICS

The magnifications produced by a convex lens for two different positions of an object are m

1

and m

2

respectively (m

1

> m

2

). If d is the distance of separation between the two positions of the object then the focal length of the lens is

HARD

Help best friend

Study later

ANSWER

Given: The magnifications produced by a convex lens for two different positions of an object are m

1

and m

2

respectively (m

1

>m

2

). If d is the distance of separation between the two positions of the object

To FIND the focal length of the lens

Solution:

Separation between object and image for first position,

D=v+u

where u,v are the object distance and image distance, respectively.

So, magnification

m

1

=

u

v

⟹v=um

1

....(i)

When the lens is at second position,

d=v-u

So, m

2

=

v

u

⟹u=vm

2

......(II)

So, m

1

m

2

=1

From lens equation and using eqn (i), we get

f

1

=

v

1

+

u

1

⟹

f

1

=

um

1

+

u

1

⟹

f

1

=

um

1

1+m

1

⟹u=

m

1

f(m

1

+1)

.........(iii)

From lens equation and using eqn (ii), we get

f

1

=

v

1

+

u

1

⟹

f

1

=

vm

2

1

+

v

1

⟹

f

1

=

vm

2

1+m

2

⟹v=

m

2

f(m

2

+1)

.........(iv)

substituting the values of eqn(iii) and(iv) in the following equation, we get

d=v−u

⟹d=

m

2

f(m

2

+1)

−

m

1

f(m

1

+1)

⟹d=f(

m

1

m

2

m

1

(m

2

+1)−m

2

(m

1

+1)

)

⟹f=

m

1

m

2

+m

1

−m

1

m

2

−m

2

DM

1

m

2

⟹f=

m

1

−m

2

dm

1

m

2

but m

1

m

2

=1

Therefore the focal length becomes,

f=

m

1

−m

2

d

solution

Discussion

2.	Discuss the advantage and two limitations of using dimensional method.
Answer» align="absmiddle" alt="\bold{\huge{\fbox{\color{blue} {Answers}}}}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cfbox%7B%5Ccolor%7Bblue%7D%20%7BAnswers%7D%7D%7D%7D" title="\bold{\huge{\fbox{\color{blue} {Answers}}}}"> Defining dimentionless constants or units that wouldmake the physical EFFECT to be studied easier. When a dimentionless quantity is formulated, the COMBINATION of experiments required for describing in them will be less when compared to the large NUMBER of participant variables in correlation. Explanation: please Make As Brainlist Answers

2.

Discuss the advantage and two limitations of using dimensional method.

Answer»

align="absmiddle" alt="\bold{\huge{\fbox{\color{blue} {Answers}}}}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cfbox%7B%5Ccolor%7Bblue%7D%20%7BAnswers%7D%7D%7D%7D" title="\bold{\huge{\fbox{\color{blue} {Answers}}}}">

Defining dimentionless constants or units that wouldmake the physical EFFECT to be studied easier. When a dimentionless quantity is formulated, the COMBINATION of experiments required for describing in them will be less when compared to the large NUMBER of participant variables in correlation.

Explanation:

please Make As Brainlist Answers

Discussion

3.	mary ran 6km south, turned north, and continued running she did not keep track of how far she ran north, but she stopped when she arrived at a shop she knows is exactly 10 km from her starting point
Answer» she RAN for 6+6+10=22km (in SOUTH 6km, return to STARTING POINT 6km CONTINUE running to 10km)

3.

mary ran 6km south, turned north, and continued running she did not keep track of how far she ran north, but she stopped when she arrived at a shop she knows is exactly 10 km from her starting point

Answer»

she RAN for 6+6+10=22km

(in SOUTH 6km, return to STARTING POINT 6km CONTINUE running to 10km)

Discussion

4.	Khad padarth ko Parikshit karne ke liye kaun se adhatu ka prayog Kiya jata hai
Answer» ong>Answer: HI *mate* *here* is *your* *answer* *NITROGEN* IS *USED* IN *FOOD* *PRESERVATION* *PROCESS. HOPE* IT *HELPED* *YOU* ✌ *MARK* ME AS *BRAINLIST* ✌

4.

Khad padarth ko Parikshit karne ke liye kaun se adhatu ka prayog Kiya jata hai

Answer»

ong>Answer:

HI mate

here is your answer

NITROGEN IS USED IN FOOD PRESERVATION PROCESS.

HOPE IT HELPED YOU ✌

MARK ME AS BRAINLIST ✌

Discussion

5.	Scalars and vectors
Answer» ong>Explanation: quantity is EITHER a vector or a scalar. These TWO categories can be distinguished from one another by their distinct definitions: SCALARS are quantities that are FULLY described by a magnitude (or numerical value) alone. Vectors are quantities that are fully described by both a magnitude and a direction. please make me brainlist Dear and FOLLOW me

5.

Scalars and vectors

Answer»

ong>Explanation:

quantity is EITHER a vector or a scalar. These TWO categories can be distinguished from one another by their distinct definitions: SCALARS are quantities that are FULLY described by a magnitude (or numerical value) alone. Vectors are quantities that are fully described by both a magnitude and a direction.

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Discussion

6.	What is the capital of jharkhand,goa ,bihar
Answer» >JHARKHAND- RANCHI GOA- PANAJI BIHAR- PATNA HOPE ITS HELP. PLZ MARK ME AS BRAINLIEST. FOLLOW @pubgqueen

6.

What is the capital of jharkhand,goa ,bihar

Answer»

>JHARKHAND- RANCHI

GOA- PANAJI

BIHAR- PATNA

HOPE ITS HELP.

PLZ MARK ME AS BRAINLIEST.

FOLLOW @pubgqueen

Discussion

7.	explain the effect of increasing the magnitude of gate current and junction temperature on forward and reverse leakage current?
Answer» ONG>EXPLANATION: this is your ANSWER BRO

Discussion

8.	Hiidefine 1ly and 1 parsec
Answer» ONG>Explanation: 1ly=9.46×10 to the power 15 mtr. Parsec=parsec is the distance from where the semi major axis of the earth subtends an ANGLE of 1 second. 1 parsec=3.08×10 to the power 16. ASTRONOMICAL unit=1 astronomical unit is equal to the mean distance between earth and sun. please MAKE me brainlist Dear and FOLLOW me

8.

Hiidefine 1ly and 1 parsec

Answer» ONG>Explanation:

1ly=9.46×10 to the power 15 mtr. Parsec=parsec is the distance from where the semi major axis of the earth subtends an ANGLE of 1 second. 1 parsec=3.08×10 to the power 16. ASTRONOMICAL unit=1 astronomical unit is equal to the mean distance between earth and sun.

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Discussion

9.	PHYSICS CHAPTER 2A. Objective Questions 1. Write true or false for each statement(a) Equal volumes of the two different substances have equal masses.(b) The density of a piece of brass will change by changing its size or shape.(c) The density of a liquid decreases with increase in its temperature.(d) Relative density of water is 1.0.(e) Relative density of a substance is expressed in gm-3.(f) When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the volume of the liquid displaced by it.(g) A body experiences the same buoyant force while floating in water or alcohol. A body experiences the same buoyant force when it floats or sinks in water.(h) A body floats in a liquid when its weight becomes equal to the weight of the liquid displaced by its submerged part(i) ) A body while floating, sinks deeper in a liquid of low density than in a liquid of high density.2. Fill in the blanks (a) 1kg is the mass of______ water at 4°C. (b) Mass = density × ______ (c) The S.I. unit of density is _____ (d) Density of water is ______ (e) 1 g cm-3 = ______(f) The density of a body which sinks in water is ____than 1000 Kg (g) A body sinks in a liquid A; but floats in a liquid. The density of liquid is ____than the density of liquid B (h) A body sinks in water, but a body Y floats on water. The density of the body is ____ than the density of the body. (i) The buoyant force experienced by a body when floating in salt-water is ______ to or same that of when floating in pure water. (j) The weight of a body floating in a liquid is _____3. . Select the correct alternative (a) The correct relation is 1. Density = Mass × Volume 2. Mass = Density × Volume 3. Volume = Density × Mass 4. Density = Mass + Volume (b) The relative density of alcohol is 0.8 its density is 1. 0.8 2. 800 kg 3 3. 800 g 34. 0.8 kg −3(c) A block of wood of density 0.8gcm-3 has a volume of 60cm3 . The mass of block is 1. 1. 60.8 g 2. 75 g 3. 48 g 4. 0.013 g(d) The density of aluminium is 2.7 g3 and that of brass 8.4 3 . The correct statement is 1. Equal masses of aluminium and brass have equal volumes 2. The mass of a certain volume of brass is more than the mass of equal volume of aluminium. 3. The volume of a certain mass of brass is more than the volume of equal mass of aluminium. 4. Equal volumes of aluminium and brass have equal masses.(e) A density bottle has a marking 25 mL on it. It means that: 1. the mass of density bottle is 25g 2. the density bottle will store 25 ml of any liquid in it 3. the density bottle will store 25 ml of water, but more volume of liquid denser than water. 4. the density bottle will store 25 ml of water, but more volume of a liquid lighter than water.(f) The correct statement is1. The buoyant force on a body is equal to the volume of the liquid displaced by it 2. The buoyant force on a body is equal to the volume of the body 3. The buoyant force on a body is equal to the weight of the liquid displaced by it 4. The buoyant force on a body is always equal to the weight of the body.(g) A piece of wood floats on water. The buoyant force on wood will be 1. zero 2. more than the weight of the wood piece 3. equal to the weight of the wood piece 4. less than the weight of the wood piece.(h) The weight of a body is more than the buoyant force experienced by it, due to a liquid. The body will 1. Sink 2. float with its some part outside the liquid 3. float just below the surface of liquid 4. Float with whole of its volume above the surface of liquid.
Answer» ONG>EXPLANATION: true true false false true false

9.

PHYSICS CHAPTER 2A. Objective Questions 1. Write true or false for each statement(a) Equal volumes of the two different substances have equal masses.(b) The density of a piece of brass will change by changing its size or shape.(c) The density of a liquid decreases with increase in its temperature.(d) Relative density of water is 1.0.(e) Relative density of a substance is expressed in gm-3.(f) When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the volume of the liquid displaced by it.(g) A body experiences the same buoyant force while floating in water or alcohol. A body experiences the same buoyant force when it floats or sinks in water.(h) A body floats in a liquid when its weight becomes equal to the weight of the liquid displaced by its submerged part(i) ) A body while floating, sinks deeper in a liquid of low density than in a liquid of high density.2. Fill in the blanks (a) 1kg is the mass of__ water at 4°C. (b) Mass = density × (c) The S.I. unit of density is _ (d) Density of water is (e) 1 g cm-3 = (f) The density of a body which sinks in water is than 1000 Kg (g) A body sinks in a liquid A; but floats in a liquid. The density of liquid is than the density of liquid B (h) A body sinks in water, but a body Y floats on water. The density of the body is than the density of the body. (i) The buoyant force experienced by a body when floating in salt-water is to or same that of when floating in pure water. (j) The weight of a body floating in a liquid is _3. . Select the correct alternative (a) The correct relation is 1. Density = Mass × Volume 2. Mass = Density × Volume 3. Volume = Density × Mass 4. Density = Mass + Volume (b) The relative density of alcohol is 0.8 its density is 1. 0.8 2. 800 kg 3 3. 800 g 34. 0.8 kg −3(c) A block of wood of density 0.8gcm-3 has a volume of 60cm3 . The mass of block is 1. 1. 60.8 g 2. 75 g 3. 48 g 4. 0.013 g(d) The density of aluminium is 2.7 g3 and that of brass 8.4 3 . The correct statement is 1. Equal masses of aluminium and brass have equal volumes 2. The mass of a certain volume of brass is more than the mass of equal volume of aluminium. 3. The volume of a certain mass of brass is more than the volume of equal mass of aluminium. 4. Equal volumes of aluminium and brass have equal masses.(e) A density bottle has a marking 25 mL on it. It means that: 1. the mass of density bottle is 25g 2. the density bottle will store 25 ml of any liquid in it 3. the density bottle will store 25 ml of water, but more volume of liquid denser than water. 4. the density bottle will store 25 ml of water, but more volume of a liquid lighter than water.(f) The correct statement is1. The buoyant force on a body is equal to the volume of the liquid displaced by it 2. The buoyant force on a body is equal to the volume of the body 3. The buoyant force on a body is equal to the weight of the liquid displaced by it 4. The buoyant force on a body is always equal to the weight of the body.(g) A piece of wood floats on water. The buoyant force on wood will be 1. zero 2. more than the weight of the wood piece 3. equal to the weight of the wood piece 4. less than the weight of the wood piece.(h) The weight of a body is more than the buoyant force experienced by it, due to a liquid. The body will 1. Sink 2. float with its some part outside the liquid 3. float just below the surface of liquid 4. Float with whole of its volume above the surface of liquid.

Answer» ONG>EXPLANATION:

true

false

true

false

Discussion

10.	. A spring loaded governor of Wilson-Hartnell type is shown in fig. Two balls each of mass 6 kg are connected by two spring. A speed of the governor is 600 r.p.m. in its mean position. Theradius of the governor ball is 150 mm. The tension in each of the spring A is 1300 N. Find the tension in the spring B for this position. Find the necessary stiffness of the spring B if the speed is 630 r.p.m. when the sleeve moves up 20 mm from mean position and if stiffness of each spring A is 10kN/m
Answer» ONG>ANSWER: Don't KNOW Sorry And it is not of my CLASS

10.

. A spring loaded governor of Wilson-Hartnell type is shown in fig. Two balls each of mass 6 kg are connected by two spring. A speed of the governor is 600 r.p.m. in its mean position. Theradius of the governor ball is 150 mm. The tension in each of the spring A is 1300 N. Find the tension in the spring B for this position. Find the necessary stiffness of the spring B if the speed is 630 r.p.m. when the sleeve moves up 20 mm from mean position and if stiffness of each spring A is 10kN/m

Answer» ONG>ANSWER:

Don't KNOW

Sorry

And it is not of my CLASS

Discussion

11.	(i)Derive the following expression of effects of partial balancing in two cylinder locomotive engine (i) Variation of attractive force (ii) Swaying couple (iii) Hammer blow(i)State the methods of force balancing of linkages by lowen and berk method.
Answer» ate the methods of force balancing of linkages by lowen and berk method. =(ans in 1st attachment) Q: Variation of ATTRACTIVE force = Attractive and repulsive forces between atoms: My UNDERSTANDING is that, as two HYDROGEN atoms APPROACH, attraction is stronger initially but repulsion is stronger at shorter distances. Taking the simplistic view that protons and electrons are discrete particles, it seems to me that, no matter how closely the atoms approach, there is a always a configuration in which the proton-electron distances can be smaller than the proton-proton and electron-electron distances. Given this, why should repulsion be stronger than attraction at short distances?

11.

(i)Derive the following expression of effects of partial balancing in two cylinder locomotive engine (i) Variation of attractive force (ii) Swaying couple (iii) Hammer blow(i)State the methods of force balancing of linkages by lowen and berk method.

Answer»

ate the methods of force balancing of linkages by lowen and berk method.

=(ans in 1st attachment)

Q: Variation of ATTRACTIVE force

=

Attractive and repulsive forces between atoms:

My UNDERSTANDING is that, as two HYDROGEN atoms APPROACH, attraction is stronger initially but repulsion is stronger at shorter distances. Taking the simplistic view that protons and electrons are discrete particles, it seems to me that, no matter how closely the atoms approach, there is a always a configuration in which the proton-electron distances can be smaller than the proton-proton and electron-electron distances. Given this, why should repulsion be stronger than attraction at short distances?

Discussion

12.	Who is the most likely to complete skill-related fitness activities?
Answer» ong>Explanation: INACTIVE individuals COMPETITIVE athletes moderately-fit athletes beginning athletes. please MAKE me brainlist DEAR and follow me

12.

Who is the most likely to complete skill-related fitness activities?

Answer»

ong>Explanation:

INACTIVE individuals COMPETITIVE athletes moderately-fit athletes beginning athletes.

please MAKE me brainlist DEAR and follow me

Discussion

13.	A kangaroo is capable of jumping to height of 2.6m . Determine the take off speed of the kangaroo
Answer» ong>Answer: 7.17 m/s Explanation: MAXIMUM HEIGHT reached by Kangaroo H=2.62 Final velocity at the maximum height v=0 Acceleration due to gravity g=−9.8 m/s 2 (minus sign for downward DIRECTION) Using v 2 −u 2 =2gH ∴ 0−u 2 =2(−9.8)(2.62) ⟹ u= 2(9.8)(2.62) =7.17 m/s

13.

A kangaroo is capable of jumping to height of 2.6m . Determine the take off speed of the kangaroo

Answer»

ong>Answer:

7.17 m/s

Explanation:

MAXIMUM HEIGHT reached by Kangaroo H=2.62

Final velocity at the maximum height v=0

Acceleration due to gravity g=−9.8 m/s

2

(minus sign for downward DIRECTION)

Using v

2

−u

2

=2gH

∴ 0−u

2

=2(−9.8)(2.62)

⟹ u=

2(9.8)(2.62)

=7.17 m/s

Discussion

14.	Robert is dribbling the basketball up the court while his teammate, Chris, runs alongside him. When a defender tries to stop Robert, he passes the ball to Chris, who shoots and scores. Which strategy are Robert and Chris displaying as they play the game?
Answer» ong>Explanation: Robert is dribbling the BASKETBALL up the COURT while his teammate, Chris, runs alongside him. When a defender TRIES to stop Robert, he passes the ball to Chris, who shoots and scores. please MAKE me brainlist Dear and FOLLOW me

14.

Robert is dribbling the basketball up the court while his teammate, Chris, runs alongside him. When a defender tries to stop Robert, he passes the ball to Chris, who shoots and scores. Which strategy are Robert and Chris displaying as they play the game?

Answer»

ong>Explanation:

Robert is dribbling the BASKETBALL up the COURT while his teammate, Chris, runs alongside him. When a defender TRIES to stop Robert, he passes the ball to Chris, who shoots and scores.

please MAKE me brainlist Dear and FOLLOW me

Discussion

15.	Sportsmanship means following the rules of the game and respecting the judgment of referees and officials. What is another example of sportsmanship?
Answer» ong>Explanation: Another example of sportsmanship is to submit to the sport or EVENT that you lose. You have to accept that despite everything, you will not always win the game and you have to accept that reality. PLEASE MAKE me brainlist Dear and FOLLOW me on

15.

Sportsmanship means following the rules of the game and respecting the judgment of referees and officials. What is another example of sportsmanship?

Answer»

ong>Explanation:

Another example of sportsmanship is to submit to the sport or EVENT that you lose. You have to accept that despite everything, you will not always win the game and you have to accept that reality.

PLEASE MAKE me brainlist Dear and FOLLOW me on

Discussion

16.	A cantilever shaft 10 cm diameter and 100 cm long is fixed at one end and other end carries a flywheel of mass 80 kg. The young’s modulus for the shaft material is 2 x 106 kg/cm2. Determine the natural frequency of longitudinal and transverse vibration of the shaft?
Answer» ong>EXPLANATION: TAKE as reference please make me BRAINLIST DEAR and follow me

16.

A cantilever shaft 10 cm diameter and 100 cm long is fixed at one end and other end carries a flywheel of mass 80 kg. The young’s modulus for the shaft material is 2 x 106 kg/cm2. Determine the natural frequency of longitudinal and transverse vibration of the shaft?

Answer»

ong>EXPLANATION:

TAKE as reference

please make me BRAINLIST DEAR and follow me

Discussion

17.	A cantilever shaft 10 cm diameter and 100 cm long is fixed at one end and other end carries a flywheel of mass 80 kg. The young’s modulus for the shaft material is 2 x 106kg/cm2. Determine the natural frequency of longitudinal and transverse vibration of the shaft
Answer» ONG>EXPLANATION: take as a reference and slove the above QUESTION PLEASE MAKE me brainlist Dear and follow me

17.

A cantilever shaft 10 cm diameter and 100 cm long is fixed at one end and other end carries a flywheel of mass 80 kg. The young’s modulus for the shaft material is 2 x 106kg/cm2. Determine the natural frequency of longitudinal and transverse vibration of the shaft

Answer» ONG>EXPLANATION:

take as a reference

and slove the above QUESTION

PLEASE MAKE me brainlist Dear and follow me

Discussion

18.	Rank the following objects from least inertia to most inertia. Baseball Small car Truck Feather Large train
Answer» ong>ANSWER: Feather, BASEBALL, Small CAR, Truck, LARGE TRain Explanation:

18.

Rank the following objects from least inertia to most inertia. Baseball Small car Truck Feather Large train

Answer»

ong>ANSWER:

Feather, BASEBALL, Small CAR, Truck, LARGE TRain

Explanation:

Discussion

19.	Light enters from the air to glass having refractive index 1.50 calculate the speed of light in the glass given :: the speed of light in vacuum is = 3' 10 8' m/s
Answer» >Answer:- $\red{\bigstar}$ Speed of light in GLASS $\large\leadsto\boxed{\sf \purple{2 × 10^8 \: m/s}}$ • Given:- Refractive INDEX of glass = 1.50 Speed of light in VACCUM = $\bf{3 \times 10^8}$ • To Find:- Speed of light in glass = ? • Solution:- We know, $\pink{\bigstar}$ $\boxed{\sf \red{Refractive \: index = \dfrac{Speed \: of \: light \: in \: vaccum}{Speed \: of \: light \: in \: medium}}}$ Hence, → $\sf{1.50 = \dfrac{3 \times 10^8}{Speed \: of \: light \: in \: glass}}$ → $\sf{Speed \: of \: light \: in \: glass = \dfrac{3 \times 10^8}{1.50}}$ → $\sf{Speed \: of \: light \: in \: glass = 2 \times 10^8}$ Therefore, the speed of light in glass is $\bf \pink{2 \times 10^8 \: m/s}$ .

19.

Light enters from the air to glass having refractive index 1.50 calculate the speed of light in the glass given :: the speed of light in vacuum is = 3' 10 8' m/s

Answer»

>Answer:-

$\red{\bigstar}$ Speed of light in GLASS

$\large\leadsto\boxed{\sf \purple{2 × 10^8 \: m/s}}$

• Given:-

Refractive INDEX of glass = 1.50

Speed of light in VACCUM = $\bf{3 \times 10^8}$

• To Find:-

Speed of light in glass = ?

• Solution:-

We know,

$\pink{\bigstar}$

$\boxed{\sf \red{Refractive \: index = \dfrac{Speed \: of \: light \: in \: vaccum}{Speed \: of \: light \: in \: medium}}}$

Hence,

→ $\sf{1.50 = \dfrac{3 \times 10^8}{Speed \: of \: light \: in \: glass}}$

→ $\sf{Speed \: of \: light \: in \: glass = \dfrac{3 \times 10^8}{1.50}}$

→ $\sf{Speed \: of \: light \: in \: glass = 2 \times 10^8}$

Therefore, the speed of light in glass is $\bf \pink{2 \times 10^8 \: m/s}$ .

Discussion

20.	A vertical double-acting steam engine develops 75 KW at 250 r.p.m. The maximum fluctuation of energy is 30% of the work done per cycle. The maximum and minimum speeds are not to vary more than 1% on either side of the mean speed. Find the mass of the flywheel required if the radius of gyration is 0.6metres.
Answer» ONG>Answer: 3456 Kg Explanation: I have attached a Handwritten Calculation to support the Answer. HOPE its OKAY.

20.

A vertical double-acting steam engine develops 75 KW at 250 r.p.m. The maximum fluctuation of energy is 30% of the work done per cycle. The maximum and minimum speeds are not to vary more than 1% on either side of the mean speed. Find the mass of the flywheel required if the radius of gyration is 0.6metres.

Answer» ONG>Answer:

3456 Kg

Explanation:

I have attached a Handwritten Calculation to support the Answer. HOPE its OKAY.

Discussion

21.	The velocity of a body moving in a straight line with a uniform acceleration of 2 m/s2 is 10 m/s. Its velocity (inm/s) after an interval of 4 sis n m/s. Find the value of n.Answer:0123 4 5 6 7 78 • 9012 34 5 6 78.9Select your answer from radio buttons.
Answer» ONG>Answer: The answer is 18m/s, i dont KNOW aboit the radio buttons BIT the answer is 18 Explanation: we have that the velocity of the moving BODY of acceleration 2m/s2 is 10 m/s then after 4 sec the velocity will be v=u+at i.e v= 10 + 4×2= 18m/s HOPE IT HELPS....

21.

The velocity of a body moving in a straight line with a uniform acceleration of 2 m/s2 is 10 m/s. Its velocity (inm/s) after an interval of 4 sis n m/s. Find the value of n.Answer:0123 4 5 6 7 78 • 9012 34 5 6 78.9Select your answer from radio buttons.

Answer» ONG>Answer:

The answer is 18m/s, i dont KNOW aboit the radio buttons BIT the answer is 18

Explanation:

we have that the velocity of the moving BODY of acceleration 2m/s2 is 10 m/s then after 4 sec the velocity will be

v=u+at

i.e v= 10 + 4×2= 18m/s

HOPE IT HELPS....

Discussion

22.	If AB vector =AB then find angle between A vector and B vector
Answer» ong>Answer: angle between VECTOR A and B is 45°

22.

If AB vector =AB then find angle between A vector and B vector

Answer»

ong>Answer:

angle between VECTOR A and B is 45°

Discussion

23.	........ energy is stored in stretched spring
Answer» ONG>Answer: Elastic POTENTIAL energy is stored in stretched spring. PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME.

23.

........ energy is stored in stretched spring

Answer» ONG>Answer:

Elastic POTENTIAL energy is stored in stretched spring.

PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME.

Discussion

24.	Vector a =i+j-2kand vector b=3i+3j-6k
Answer» ONG>Answer: PLEASE WRITE the complete QUESTION, I think you've missed some portion of it.

24.

Vector a =i+j-2kand vector b=3i+3j-6k

Answer» ONG>Answer:

PLEASE WRITE the complete QUESTION, I think you've missed some portion of it.

Discussion

25.	⦁ A box has dimensions 20 cm × 40 cm × 60 cm, and a mass of 50 kg. ⦁ Calculate the weight of the box
Answer» ong>Answer: 500 kilograms Explanation: weight = MASS X GRAVITY = 50 × 10 = 500

25.

⦁ A box has dimensions 20 cm × 40 cm × 60 cm, and a mass of 50 kg. ⦁ Calculate the weight of the box

Answer»

ong>Answer:

500 kilograms

Explanation:

weight = MASS X GRAVITY

= 50 × 10

= 500

Discussion

26.	a cricket ball of mass 0.15kg is moving with a velocity of 1.2m/s. Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.18s?
Answer» ONG>EXPLANATION: HOPE this will be HELPFUL for you

Discussion

27.	3 515. If R = 2 cal. mol-1.°C-1 and hydrogen is assumed tobe an ideal gas, specific heat of that gas at constantpressure will beA) 7 cal.g-1.°C B) 5 cal.g 1.°C-1C)3.5 cal.g-1.°C-1 D)1.25 cal.g1.°C-1
Answer» ONG>Answer: SORRY don't KNOW the truth Explanation: I hope you got it make me brainlist

27.

3 515. If R = 2 cal. mol-1.°C-1 and hydrogen is assumed tobe an ideal gas, specific heat of that gas at constantpressure will beA) 7 cal.g-1.°C B) 5 cal.g 1.°C-1C)3.5 cal.g-1.°C-1 D)1.25 cal.g1.°C-1

Answer» ONG>Answer:

SORRY don't KNOW the truth

Explanation:

I hope you got it make me brainlist

Discussion

28.	A. Objective Questions 1. Write true or false for each statement(a) A coolie does no work against the force of gravity while carrying a luggage on a road(b) The energy stored in water of a dam is the kinetic energy.(c) The energy of a flying kite is kinetic energy.(d) Work done by a boy depends on the time in which he does work.(e) Power spent by a body depends on the time for which it does work.2. Fill in the blanks:(a) Work is said to be done by a forte only when _________(b) Work done = Force x _________(c) The energy of a body is its capacity to do ________(d) The S.I. unit of energy is _______(e) The potential energy is due to its _________ n and kinetic energy of body is due to its _________(f) Gravitational potential energy U = mass × force of gravity on unit mass × ________(g) Kinetic energy = ½ × mass × _______(h) Power P = ______ /time taken(i) The S. I. unit of power is ________(j) I.H.P. = _______Select the correct alternative (a) The S.I. unit of work is 1. second 2. metre 3. joule 4. Newton(b) No work is done by a force if the body 1. moves in direction of force 2. does not move 3. moves in opposite direction 4. none of the these(c) Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is 1. same by both coolies 2. is more by coolie A than by B 3. is less by coolie A than by B 4. nothing can be said.(d) The expression of power P is 1. P=mgh 2. P=P=1/223. P=F×d 4. P=F×d/t(e) 1 H.P is equal to 1. 1 W 2. 1 J 3. 746 J 4. 746 W(f) When a boy doubles his speed, his kinetic energy becomes 1. half 2. Double3. four times 4. no change(g) A boy lifts a luggage from height 2 m to 4 m. The potential energy will become1. half2. double 3. one-third 4. one-fourth
Answer» ONG>Explanation: 1 (a) True (B) False (C) True (d) False (E) True Hope it helps you!!

28.

A. Objective Questions 1. Write true or false for each statement(a) A coolie does no work against the force of gravity while carrying a luggage on a road(b) The energy stored in water of a dam is the kinetic energy.(c) The energy of a flying kite is kinetic energy.(d) Work done by a boy depends on the time in which he does work.(e) Power spent by a body depends on the time for which it does work.2. Fill in the blanks:(a) Work is said to be done by a forte only when _____(b) Work done = Force x _(c) The energy of a body is its capacity to do (d) The S.I. unit of energy is _(e) The potential energy is due to its ___ n and kinetic energy of body is due to its _(f) Gravitational potential energy U = mass × force of gravity on unit mass × (g) Kinetic energy = ½ × mass × _(h) Power P = /time taken(i) The S. I. unit of power is (j) I.H.P. = ___Select the correct alternative (a) The S.I. unit of work is 1. second 2. metre 3. joule 4. Newton(b) No work is done by a force if the body 1. moves in direction of force 2. does not move 3. moves in opposite direction 4. none of the these(c) Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is 1. same by both coolies 2. is more by coolie A than by B 3. is less by coolie A than by B 4. nothing can be said.(d) The expression of power P is 1. P=mgh 2. P=P=1/223. P=F×d 4. P=F×d/t(e) 1 H.P is equal to 1. 1 W 2. 1 J 3. 746 J 4. 746 W(f) When a boy doubles his speed, his kinetic energy becomes 1. half 2. Double3. four times 4. no change(g) A boy lifts a luggage from height 2 m to 4 m. The potential energy will become1. half2. double 3. one-third 4. one-fourth

Answer» ONG>Explanation:

1 (a) True

(B) False

(C) True

(d) False

(E) True

Hope it helps you!!

Discussion

29.	a cricket ball of mass 0.15kg is moving with a velocity of 1.2m/s. Find the impulse on the ball and force applied by the player if he is able to stop the ball in 0.18s?
Answer» ONG>ANSWER: HOPE this will be HELPFUL for you

Discussion

30.	A uniform rod AB is resting on two supports which are at distnaces 3 m and 4 m from the ends A and B respectively. Masses of 10 kg and 8 kg are suspended from the ends A and B respectively. The weight of the rod is 45 kg and its length is 12 m. Find the reaction at the supports.
Answer» ONG>EXPLANATION: tefwgwggwwtwttwtwtwtgwefefwww ebgegwwg heheggegegge ebebvrvevrv nrbrbbrbb rbbrbdbbd hxygwhdyhw

30.

A uniform rod AB is resting on two supports which are at distnaces 3 m and 4 m from the ends A and B respectively. Masses of 10 kg and 8 kg are suspended from the ends A and B respectively. The weight of the rod is 45 kg and its length is 12 m. Find the reaction at the supports.

Answer» ONG>EXPLANATION:

tefwgwggwwtwttwtwtwtgwefefwww ebgegwwg

heheggegegge

ebebvrvevrv

nrbrbbrbb

rbbrbdbbd

hxygwhdyhw

Discussion

31.	.Find the value of 1 watt on a system in which unit of mass is 1g, unit of length is 1cm and unit of times is 1sec. Also give the name of new unit.
Answer» the ANSWER cuz can't read it so PLS type properly...just a SUGGESTION!!!

Discussion

32.	Timmy runs 500m to the left in 58.2 s. What is his average velocity?
Answer» ONG>Answer: 500 ÷ 58.2 = 8.59106529 Explanation: Speed equals DISTANCE TRAVELLED DIVIDED by time elapsed

32.

Timmy runs 500m to the left in 58.2 s. What is his average velocity?

Answer» ONG>Answer:

500 ÷ 58.2 = 8.59106529

Explanation:

Speed equals DISTANCE TRAVELLED DIVIDED by time elapsed

Discussion

33.	state snell's law of refraction of light write an expression for the absolute refractive index of a medium in terms of speed of light
Answer» ong>Answer: Snell's laws of refraction of light : (1) : The incident ray, refracted ray and NORMAL to the refracted surface LIE in same plane. (2) : Ratio of sine of ANGLE of INCIDENCE in medium 1 to that of angle of refraction in medium 2 is equal to the ratio of refractive index of medium 2 to that of 1.

33.

state snell's law of refraction of light write an expression for the absolute refractive index of a medium in terms of speed of light

Answer»

ong>Answer:

Snell's laws of refraction of light : (1) : The incident ray, refracted ray and NORMAL to the refracted surface LIE in same plane. (2) : Ratio of sine of ANGLE of INCIDENCE in medium 1 to that of angle of refraction in medium 2 is equal to the ratio of refractive index of medium 2 to that of 1.

Discussion

34.	A rock is dropped from a high bridge. At the end of 3 seconds of free fall the speed of the rock is (in centimeters per second)
Answer» 3=2940 cm/s Hope it HELPS Have a NICE DAY MATE

34.

A rock is dropped from a high bridge. At the end of 3 seconds of free fall the speed of the rock is (in centimeters per second)

Answer»

3=2940 cm/s

Hope it HELPS

Have a NICE DAY MATE

Discussion

35.	State and explain newton's second law of motion in about 10 para it is long answer type question
Answer» ONG>Explanation: Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. ... So, the force/mass ratio (from the equation acceleration = force/mass) is the same for each. Therefore, the acceleration is the same and they reach the GROUND at the same time. MOREOVER Second law :- In an inertial FRAME of reference, the vector sum of the FORCES F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. (It is assumed here that the mass m is constant – see below.)

35.

State and explain newton's second law of motion in about 10 para it is long answer type question

Answer» ONG>Explanation:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. ... So, the force/mass ratio (from the equation acceleration = force/mass) is the same for each. Therefore, the acceleration is the same and they reach the GROUND at the same time.

MOREOVER

Second law :- In an inertial FRAME of reference, the vector sum of the FORCES F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. (It is assumed here that the mass m is constant – see below.)

Discussion

36.	When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, -1/2. Where should the object be placed to get a magnification of, -1/5?
Answer» ONG>ANSWER: answer is o is a HALF so answer o

36.

When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, -1/2. Where should the object be placed to get a magnification of, -1/5?

Answer» ONG>ANSWER:

answer is

o is a HALF so answer o

Discussion

37.	A body w moving awardingly, to the equation x = whereasbands are constants. The omeleradio of bodya6) ab + bct c] Jab26-6t d]36-6ctis.at abt2
Answer» ONG>Answer: A Body Is Moving According To The Equation x=at+bt2−ct3Where x= DISPLACEMENT a, B, c Are Constants.

Discussion

38.	Which describes a feature of the physical properties of all substances? can dissolve in water can conduct heat and electricity rearranges atoms to form new substances does not change the identity of a substance
Answer» not CHANGE the IDENTITY of SUBSTANCE

Discussion

39.	EXPLAIN FLEMING'S LEFT HAND RULE?class 10thh
Answer» $\huge\bf \red \: \boxed{\mathfrak{ Answer\: ヅ}}$ FLEMING'S LEFT HAND RULE:- ✔️According to this rule, stretch the thumb, FOREFINGER and middle finger of your left hand such that they are mutually perpendicular. ✔️If the forefinger POINTS in the direction of magnetic field and the middle finger in the direction of current, then the thumb will POINT in the direction of motion or the force acting on the CONDUCTOR.

39.

EXPLAIN FLEMING'S LEFT HAND RULE?class 10thh

Answer»

$\huge\bf \red \: \boxed{\mathfrak{ Answer\: ヅ}}$

FLEMING'S LEFT HAND RULE:-

✔️According to this rule, stretch the thumb, FOREFINGER and middle finger of your left hand such that they are mutually perpendicular.

✔️If the forefinger POINTS in the direction of magnetic field and the middle finger in the direction of current, then the thumb will POINT in the direction of motion or the force acting on the CONDUCTOR.

Discussion

40.	5.An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?
Answer» ong>Explanation: Given Mass of the automobile vehicle, m= 1500 kg Final velocity, v= 0 (FINALLY the automobile stops) Acceleration of the automobile, a= 1.7 ms2 Find out We need to find the force between the vehicle and ROAD if the vehicle is to be stopped with a NEGATIVE acceleration of 1.7 ms2 Solution Using, Newton’s second law of motion Formula Force = Mass x Acceleration Substituting the known values we get F= 1500 x (1.7) = 2550 N Answer Hence, the force between the automobile and the road is 2550 N, in the direction opposite to the motion of the automobile.

40.

5.An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?

Answer»

ong>Explanation:

Given

Mass of the automobile vehicle, m= 1500 kg

Final velocity, v= 0 (FINALLY the automobile stops)

Acceleration of the automobile, a= 1.7 ms2

Find out

We need to find the force between the vehicle and ROAD if the vehicle is to be stopped with a NEGATIVE acceleration of 1.7 ms2

Solution

Using, Newton’s second law of motion

Formula

Force = Mass x Acceleration

Substituting the known values we get

F= 1500 x (1.7) = 2550 N

Answer

Hence, the force between the automobile and the road is 2550 N, in the direction opposite to the motion of the automobile.

Discussion

41.	What is an ideal gas and what are its property
Answer» $\huge\bf \red \: \boxed{\mathfrak{ Answer\: ヅ}}$ ⭕️IDEAL gases are gases which obey ideal gas equation i.e [PV= nRT ] or ️gas laws under all conditions of temperature and pressure. ⭕️But unfortunately, there is no ideal gas. ⭕️ hence we can say that this ideal gas concept is just HYPOTHETICAL!!!!!!!!! ✔️✔️PV=nRT ➖P- pressure ➖V-Volume ➖n-no. of moles ➖R- Gas constant ➖T-temprature. ✔️✔️✔️Properties :- ⭕️The PARTICLES have no forces acting among them. ⭕️An ideal gas CONSISTS of a large number of identical molecules. ⭕️The molecules obey Newton's laws of motion and hence they move in random motion. ⭕️Particles do not take any space i.e their atomic volume will be always completely IGNORED!!!!! :)

41.

What is an ideal gas and what are its property

Answer»

$\huge\bf \red \: \boxed{\mathfrak{ Answer\: ヅ}}$

⭕️IDEAL gases are gases which obey ideal gas equation i.e [PV= nRT ] or ️gas laws under all conditions of temperature and pressure.

⭕️But unfortunately, there is no ideal gas.

⭕️ hence we can say that this ideal gas concept is just HYPOTHETICAL!!!!!!!!!

✔️✔️PV=nRT

➖P- pressure

➖V-Volume

➖n-no. of moles

➖R- Gas constant

➖T-temprature.

✔️✔️✔️Properties :-

⭕️The PARTICLES have no forces acting among them.

⭕️An ideal gas CONSISTS of a large number of identical molecules.

⭕️The molecules obey Newton's laws of motion and hence they move in random motion.

⭕️Particles do not take any space i.e their atomic volume will be always completely IGNORED!!!!!

:)

Discussion

42.	the moment of inertia of a uniform disc about his axis is 1/2 MR² then what will be the moment of inertia about his diameter.
Answer» ong>Answer: Let M and R be the mass and radius of the disc respectively. MOMENT of INERTIA about AB is I. Moment of inertia about an AXIS passing through O and perpendicular to the plane, I O = 2 1 MR 2 Moment of inertia about an axis passing through C and perpendicular to the plane, I C =I O +MR 2 I C =3I O Using perpendicular axis theorem, ⟹I O =2I Thus I C =6l

42.

the moment of inertia of a uniform disc about his axis is 1/2 MR² then what will be the moment of inertia about his diameter.

Answer»

ong>Answer:

Let M and R be the mass and radius of the disc respectively.

MOMENT of INERTIA about AB is I.

Moment of inertia about an AXIS passing through O and perpendicular to the plane, I

O

=

2

1

MR

2

Moment of inertia about an axis passing through C and perpendicular to the plane, I

C

=I

O

+MR

2

I

C

=3I

O

Using perpendicular axis theorem, ⟹I

O

=2I

Thus I

C

=6l

Discussion

43.	ball rolling down a hill was displace 2.5 m while uniformly accelerating from rest. If the final velocity was 5 m/s. What was the rate of acceleration?
Answer» ONG>ANSWER: dngfndbsbshshxdbmxyxnisgsy

Discussion

44.	From a sphere of radius 1 m, a sphere of radius 0.5 m is removed from the edge. The shift inthe C.M. is1) 13/16 m2) 16/13m3) 14/13 m4) 1/14 m
Answer» ong>HII MY NAME ALSO VIVEK . AND UR ANSWER IS 1 / 14 ... HOPE IT HELPS YOU

44.

From a sphere of radius 1 m, a sphere of radius 0.5 m is removed from the edge. The shift inthe C.M. is1) 13/16 m2) 16/13m3) 14/13 m4) 1/14 m

Answer»

ong>HII MY NAME ALSO VIVEK .

AND UR ANSWER IS 1 / 14 ...

HOPE IT HELPS YOU

Discussion

45.	How speed of light is high and wavelength of light is slow??
Answer» ht TRAVELS Faster in air, but slower in GLASS.In different materials, the speed of light slows down to less than C. That is because light refracts when it encounters different mediums. ... The equation is n=c/v. In water, the REFRACTIVE index is 1.3 and in glass, it is 1.5. Therefore, light travels faster through water, than it travels through glass. Follow me......

45.

How speed of light is high and wavelength of light is slow??

Answer»

ht TRAVELS Faster in air, but slower in GLASS.In different materials, the speed of light slows down to less than C. That is because light refracts when it encounters different mediums. ... The equation is n=c/v. In water, the REFRACTIVE index is 1.3 and in glass, it is 1.5. Therefore, light travels faster through water, than it travels through glass.

Follow me......

Discussion

46.	6.An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer» ωєя __________ Mass of the object, M1 = 1 kg Velocity of the object before COLLISION, v1 = 10 m/s Mass of the stationary wooden block, m2 = 5 kg Velocity of the wooden block before collision, v2 = 0 m/s ˆ´ Total momentum before collision = m1 v1 + m2 v2 = 1 (10) + 5 (0) = 10 kg m sˆ’1 It is given that after collision, the object and the wooden block stick together. Total mass of the combined SYSTEM = m1 + m2 Velocity of the combined object = v According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision $m1 v1 + m2 v2 = (m1 + m2) v \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ v = 10 / <klux>6</klux> \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ v = 10 / 6 \\ = 5 / 3$ The total momentum after collision is also 10 kg m/s. Total momentum just before the impact = 10 kg m sˆ’1 Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1 Hence, velocity of the combined object after collision = 5 / 3 ms-1

46.

6.An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer»

ωєя

__________

Mass of the object, M1 = 1 kg

Velocity of the object before COLLISION, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

ˆ´ Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m sˆ’1

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined SYSTEM = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

$m1 v1 + m2 v2 = (m1 + m2) v \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ v = 10 / <klux>6</klux> \\ [tex]m1 v1 + m2 v2 = (m1 + m2) v \\ 1 (10) + 5 (0) = (1 + 5) v \\ v = 10 / 6 \\ = 5 / 3$

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m sˆ’1

Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1

Hence, velocity of the combined
object after collision = 5 / 3 ms-1

Discussion

47.	Does wavelength is inversely proportional to velocity????
Answer» ong>Answer: f = FREQUENCY of the wave, v = VELOCITY or speed of the wave. The wavelength and frequency of a wave are INVERSELY proportional to each other both are directly proportional to the velocity of the wave. Explanation: Hope it helps you DEAR ✍️

47.

Does wavelength is inversely proportional to velocity????

Answer»

ong>Answer:

f = FREQUENCY of the wave, v = VELOCITY or speed of the wave. The wavelength and frequency of a wave are INVERSELY proportional to each other both are directly proportional to the velocity of the wave.

Explanation:

Hope it helps you DEAR ✍️

Discussion

48.	8.How many types of Inertia we can see while driving a car.
Answer» ONG>Answer: Inertia is CLASSIFIED into three types: inertia of REST, inertia of MOTION and inertia of direction. Explanation: Inertia is classified into three types: inertia of rest, inertia of motion and inertia of direction.

48.

8.How many types of Inertia we can see while driving a car.

Answer» ONG>Answer:

Inertia is CLASSIFIED into three types: inertia of REST, inertia of MOTION and inertia of direction.

Explanation:

Inertia is classified into three types: inertia of rest, inertia of motion and inertia of direction.

Discussion

49.	Find angle between the two vectors, A=5i^+j^ B=2i^+4j^
Answer» To find : Angle between vectors $\sf{\vec{A}}$ = 5 $\sf{\hat{i}}$ + $\sf{\hat{j}}$ and $\sf{\vec{B}}$ = 2 $\sf{\hat{i}}$ + 4 $\sf{\hat{j}}$ ANSWER : ▶ Dot product of $\sf{\vec{A}}$ and $\sf{\vec{B}}$ $:\implies\sf{\vec{A}\;.\;\vec{B}= (5\;\hat{i} + \hat{j} ) \;.\; ( 2\;\hat{i} + 4\;\hat{j})}$ $:\implies\sf{\vec{A}\;.\;\vec{B}= (5) (2) + (1) (4) }$ $:\implies\sf{\vec{A}\;.\;\vec{B}= 14}$ ▶ Magnitude of vectors $:\implies\sf{\| \vec{A} \| = \sqrt{ (5)^2 + (1)^2 } }$ $:\implies\sf{\| \vec{A} \| = \sqrt{26} }$ and $:\implies\sf{\|B\|=\sqrt{(2)^2+ (4)^2}}$ $:\implies\sf{\|B\|=\sqrt{20}}$ ▶ LET, Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ be $\theta$ then, $:\implies\sf{cos\;\theta=\dfrac{\vec{A} \;.\;\vec{B}}{\|\vec{A}\|\;.\;\|\vec{B}\|}}$ $:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{26}\;.\;\sqrt{20}}}$ $:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{520}}}$ $:\implies\sf{cos\;\theta=\dfrac{14}{2\sqrt{130}}}$ $:\implies\sf{cos\;\theta=\dfrac{7}{\sqrt{130}}}$ $:\implies\sf{cos\;\theta=\dfrac{7\sqrt{130}}{130}}$ $:\implies\sf{\theta=cos^{-1}\bigg(\dfrac{7\sqrt{130}}{130}\bigg)}$ $:\implies\boxed{\boxed{\large{\sf{\;\;\theta=52.125\;^{\circ}\;}}}}$ THEREFORE, Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ will be 52.125°. (APPROX.)

49.

Find angle between the two vectors, A=5i^+j^ B=2i^+4j^

Answer»

To find :

Angle between vectors $\sf{\vec{A}}$ = 5 $\sf{\hat{i}}$ + $\sf{\hat{j}}$ and $\sf{\vec{B}}$ = 2 $\sf{\hat{i}}$ + 4 $\sf{\hat{j}}$

ANSWER :

▶ Dot product of $\sf{\vec{A}}$ and $\sf{\vec{B}}$

$:\implies\sf{\vec{A}\;.\;\vec{B}= (5\;\hat{i} + \hat{j} ) \;.\; ( 2\;\hat{i} + 4\;\hat{j})}$

$:\implies\sf{\vec{A}\;.\;\vec{B}= (5) (2) + (1) (4) }$

$:\implies\sf{\vec{A}\;.\;\vec{B}= 14}$

▶ Magnitude of vectors

$:\implies\sf{| \vec{A} | = \sqrt{ (5)^2 + (1)^2 } }$

$:\implies\sf{| \vec{A} | = \sqrt{26} }$

and

$:\implies\sf{|B|=\sqrt{(2)^2+ (4)^2}}$

$:\implies\sf{|B|=\sqrt{20}}$

▶ LET, Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ be $\theta$

then,

$:\implies\sf{cos\;\theta=\dfrac{\vec{A} \;.\;\vec{B}}{|\vec{A}|\;.\;|\vec{B}|}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{26}\;.\;\sqrt{20}}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{520}}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{2\sqrt{130}}}$

$:\implies\sf{cos\;\theta=\dfrac{7}{\sqrt{130}}}$

$:\implies\sf{cos\;\theta=\dfrac{7\sqrt{130}}{130}}$

$:\implies\sf{\theta=cos^{-1}\bigg(\dfrac{7\sqrt{130}}{130}\bigg)}$

$:\implies\boxed{\boxed{\large{\sf{\;\;\theta=52.125\;^{\circ}\;}}}}$

THEREFORE,

Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ will be 52.125°. (APPROX.)

Discussion

50.	When we press the bulb of a dropper, with its nozzle kept under water, air in the dropper is seen to escape in the form of bubbles. If we then release the pressure on the bulb, water gets filled in the dropper. The rise of water in the dropper is due to - a) pressure of water in the container b) weight of the bulb c) gravity of the earthd) atmospheric pressure
Answer» ong>Answer: answer is d)atmospheric PRESSURE. since when the dropper is released there is a low pressure on the inside. DUE to this decrease or CHANGE in pressure , the outer ATMOSPHERIC pressure causes the rise in the WATER level

50.

When we press the bulb of a dropper, with its nozzle kept under water, air in the dropper is seen to escape in the form of bubbles. If we then release the pressure on the bulb, water gets filled in the dropper. The rise of water in the dropper is due to - a) pressure of water in the container b) weight of the bulb c) gravity of the earthd) atmospheric pressure

Answer»

ong>Answer:

answer is d)atmospheric PRESSURE. since when the dropper is released there is a low pressure on the inside. DUE to this decrease or CHANGE in pressure , the outer ATMOSPHERIC pressure causes the rise in the WATER level

Discussion

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