67486 + Interview Questions in Secondary School in Physics

1.	Click on the photo u can see questionplease answer all the questions
Answer» ANSWER: SIMPLE not SURE if it's WRONG

1.

Click on the photo u can see questionplease answer all the questions

Answer»

ANSWER:

SIMPLE not SURE if it's WRONG

Discussion

2.	A ray of light strikes mirror at angle of 450 what is the angle of reflection
Answer» Answer: the ANGLE of incidence and angle of REFLECTION is same so the angle of reflection is also 450

2.

A ray of light strikes mirror at angle of 450 what is the angle of reflection

Answer»

Answer:

the ANGLE of incidence and angle of REFLECTION is same so the angle of reflection is also 450

Discussion

3.	two spherical planets A and B have same mass but densities in ratio 8:1 the ratio accn due to garivty at the surface of AandB is
Answer» Answer: we know g= $\frac{GM}{r^{2} }$ M=density×volume g= $\frac{G\\4/3\pi r^3\\ <klux>BETA</klux> }{r\\}$ here beta is density now similarly do it for other divide both and put VALUE of density do it on UR own if your answer does not come ASK me

3.

two spherical planets A and B have same mass but densities in ratio 8:1 the ratio accn due to garivty at the surface of AandB is

Answer»

Answer:

we know g= $\frac{GM}{r^{2} }$

M=density×volume

g= $\frac{G\\4/3\pi *r^3\\ *<klux>BETA</klux> }{r\\}$ here beta is density

now similarly do it for other

divide both and put VALUE of density

do it on UR own if your answer does not come ASK me

Discussion

4.	Therefractiveindexofeye lens is ........... .
Answer» Answer: it is about 9 mm in DIAMETER and 4 mm thick ,the CRYSTALLINE lens provide 20 % of refracting power to eye. EXPLANATION: The index ranges from 1.406 at the centre to about 1. 386 in outer layer . hope you like this answer ....

4.

Therefractiveindexofeye lens is ........... .

Answer»

Answer:

it is about 9 mm in DIAMETER and 4 mm thick ,the CRYSTALLINE lens provide 20 % of refracting power to eye.

EXPLANATION:

The index ranges from 1.406 at the centre to about 1. 386 in outer layer .

hope you like this answer ....

Discussion

5.	At what separation should two equal charges, 1.0c each be placed so that the force between them equals the weight of a 50kg person?
Answer» Answer: 4242.6m Explanation: F=mg=50kg×10m/s^2=500N F=Kq^2/r^2 r^2=9×10^9×1×1/500 r^2=4242.6 or 4.2426×10^3

5.

At what separation should two equal charges, 1.0c each be placed so that the force between them equals the weight of a 50kg person?

Answer»

Answer:

4242.6m

Explanation:

F=mg=50kg×10m/s^2=500N

F=Kq^2/r^2

r^2=9×10^9×1×1/500

r^2=4242.6

or

4.2426×10^3

Discussion

6.	The take off speed of a Boeing 747 is 252kmph. If it can accelerate at 1.5ms-2 , what should be the minimum length of the runway?
Answer» Answer: 9hi jljkkklnpk GO lkjjuhiti

6.

The take off speed of a Boeing 747 is 252kmph. If it can accelerate at 1.5ms-2 , what should be the minimum length of the runway?

Answer»

Answer:

9hi jljkkklnpk GO lkjjuhiti

Discussion

7.	Give one example for the following type of motiona) Linear motion-b) Rotational motion-c) Periodic motion-d) Oscillatory motion-
Answer» Answer: 1) a car moving on a straight road 2) A PERSON performing a Somersault, is an example of rotational MOTION. When we open a cap of any soda bottle, it jumps up in the air because of pressure released and is an example of rotational 3)Examples of periodic motion INCLUDE a rocking chair, swing, tuning fork, and a pendulum. A satellite orbiting our planet and a mass 4)example of a pendulum

7.

Give one example for the following type of motiona) Linear motion-b) Rotational motion-c) Periodic motion-d) Oscillatory motion-

Answer»

Answer:

1) a car moving on a straight road

2) A PERSON performing a Somersault, is an example of rotational MOTION. When we open a cap of any soda bottle, it jumps up in the air because of pressure released and is an example of rotational

3)Examples of periodic motion INCLUDE a rocking chair, swing, tuning fork, and a pendulum. A satellite orbiting our planet and a mass

4)example of a pendulum

Discussion

8.	Parallel light of wavelength 6563 Å is incident normally on a slit 0.3850 mm wide. A lens with a focal length of 50 cm is located just behind the slit. Find the distance from the centre of the Principal maximum to (a) the first minima (b) the fifth minimum
Answer» ANSWER: The correct answer is OPTION (b) the fifth minimum

8.

Parallel light of wavelength 6563 Å is incident normally on a slit 0.3850 mm wide. A lens with a focal length of 50 cm is located just behind the slit. Find the distance from the centre of the Principal maximum to (a) the first minima (b) the fifth minimum

Answer»

ANSWER:

The correct answer is OPTION (b) the fifth minimum

Discussion

9.	A constant torque of 120Nm is required to keep the shaft of engine rotating at 1000rpm.calculate the output power of the engine.
Answer» Power=torque×angular velocity Frequency=1000/60=16.66 rps Angular velocity=2πf =2×3.14×16.66=104.62 Power=120×104.6 Power=12555 I HOPE this HELPS ( ╹▽╹ )

9.

A constant torque of 120Nm is required to keep the shaft of engine rotating at 1000rpm.calculate the output power of the engine.

Answer»

Power=torque×angular velocity

Frequency=1000/60=16.66 rps

Angular velocity=2πf

=2×3.14×16.66=104.62

Power=120×104.6

Power=12555

I HOPE this HELPS ( ╹▽╹ )

Discussion

10.	The magnification is greater than 1then the image obtained is .............. .
Answer» ANSWER: enlarged Explanation: HOPE 8T HELPS U

10.

The magnification is greater than 1then the image obtained is .............. .

Answer»

ANSWER:

enlarged

Explanation:

HOPE 8T HELPS U

Discussion

11.	show that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%?
Answer» Given: A free jet striking normally on a series of flat plates mounted on the periphery of a wheel. To Find: Show that the EFFICIENCY of jet can never exceed 50%. Solution: Let we take: $\mathbf{\textrm{Density of fluid }\ = \rho \ \dfrac{kg}{m^{3}}}\\$ $\mathbf{\textrm{Velocity of fluid }\ = v \ \dfrac{m}{s}}$ $\mathbf{\textrm{Speed of wheel }\ = u \ \dfrac{m}{s}}$ $\mathbf{\textrm{Exit area of nozzle }\ = a \ (m^{2})}$ Mass of fluid strike with blade is : $\mathbf{\textrm{Mass of fluid }\ = m =\rho \times a\times V \ \ \ \ \ \dfrac{kg}{s}}$ $\mathbf{\textrm{kinetic energy of fluid }\ = K.E =\dfrac{1}{2}\rho \times a\times V \times V^{2}\ \ \ \ \ (J)}$ ...1) Relative velocity of jet strike with blade with RESPECT to blade = V- u Work DONE by jet on blade: $\mathbf{\textrm{Work done }\ = W =\rho \times a\times V(V-u)\times u\ \ \ \ \ \ (J)}$ ...2) $\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{Work\ done\ by\ fluid}{Kinetic \ energy\ of\ fluid}}$ $\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{\rho \times a\times V(V-u)\times u}{\dfrac{1}{2}\rho \times a\times V \times V^{2}}}$ On simplify: $\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{2\times (V-u)\times u}{V^{2}}}$ ...3) From here, we see that efficiency of jet depend upon speed of wheel for a particular value of speed of jet: On differentiating efficiency with respect to blade speed , we get: $\mathbf{u = \dfrac{V}{2}}$ ...4) This is required condition of blade speed for maximum efficiency of jet: Putting value of u from equation 4 to equation 3): $\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=\dfrac{2\times (V-\frac{V}{2})\times \frac{V}{2}}{V^{2}}}$ On solving , we get: $\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=0.5 = 50}$ % HENCE PROVE that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%.

11.

show that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%?

Answer»

Given:

A free jet striking normally on a series of flat plates mounted on the periphery of a wheel.

To Find:

Show that the EFFICIENCY of jet can never exceed 50%.

Solution:

Let we take:

$\mathbf{\textrm{Density of fluid }\ = \rho \ \dfrac{kg}{m^{3}}}\\$

$\mathbf{\textrm{Velocity of fluid }\ = v \ \dfrac{m}{s}}$

$\mathbf{\textrm{Speed of wheel }\ = u \ \dfrac{m}{s}}$

$\mathbf{\textrm{Exit area of nozzle }\ = a \ (m^{2})}$

Mass of fluid strike with blade is :

$\mathbf{\textrm{Mass of fluid }\ = m =\rho \times a\times V \ \ \ \ \ \dfrac{kg}{s}}$

$\mathbf{\textrm{kinetic energy of fluid }\ = K.E =\dfrac{1}{2}\rho \times a\times V \times V^{2}\ \ \ \ \ (J)}$ ...1)

Relative velocity of jet strike with blade with RESPECT to blade = V- u

Work DONE by jet on blade:

$\mathbf{\textrm{Work done }\ = W =\rho \times a\times V(V-u)\times u\ \ \ \ \ \ (J)}$ ...2)

$\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{Work\ done\ by\ fluid}{Kinetic \ energy\ of\ fluid}}$

$\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{\rho \times a\times V(V-u)\times u}{\dfrac{1}{2}\rho \times a\times V \times V^{2}}}$

On simplify:

$\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{2\times (V-u)\times u}{V^{2}}}$ ...3)

From here, we see that efficiency of jet depend upon speed of wheel for a particular value of speed of jet:

On differentiating efficiency with respect to blade speed , we get:

$\mathbf{u = \dfrac{V}{2}}$ ...4)

This is required condition of blade speed for maximum efficiency of jet:

Putting value of u from equation 4 to equation 3):

$\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=\dfrac{2\times (V-\frac{V}{2})\times \frac{V}{2}}{V^{2}}}$

On solving , we get:

$\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=0.5 = 50}$ %

HENCE PROVE that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%.

Discussion

12.	Why should we avoid the use of fossil fuels as sources of energy?
Answer» Answer: because when they are used ones they take THOUSANDS to millions of years to recover. We should use RENEWABLE resources LIKE wind energy, hydroelectric power, ETC..

12.

Why should we avoid the use of fossil fuels as sources of energy?

Answer»

Answer:

because when they are used ones they take THOUSANDS to millions of years to recover. We should use RENEWABLE resources LIKE wind energy, hydroelectric power, ETC..

Discussion

13.	An iron ball and a feather are both droped from a height of 20 m. A) what are the time taken by the iron ball and feather to reach the ground. B) what are the velocities of iron ball and feather to reach the ground.
Answer» Answer: 20/20=1 is the VELOCITIES

13.

An iron ball and a feather are both droped from a height of 20 m. A) what are the time taken by the iron ball and feather to reach the ground. B) what are the velocities of iron ball and feather to reach the ground.

Answer»

Answer:

20/20=1 is the VELOCITIES

Discussion

14.	A cube of edge length 2.00m is placed in a region and which the electric field is given by E=(20.00-28.00z) k N/C. Determine the net charge contained within the cube
Answer» Given : Edge LENGTH of the given cube = 2 m The electric field in that region in which the cube is placed is given by : E= ( 20.00 - 28.00z ) $\frac{kN}{C}$ To Find : The amount of net charge contained within the cube = ? Solution: Here the electric field along x and y direction is ZERO since E is acting along only z direction . In this cube we have taken two surfaces as shown in the attached fig . So, the net electric flux = Flux from 2ND surface - flux from 1st surface ( ∴ flux going out is taken as positive and flux coming in is taken as negative) So, $\Phi_{Enet} = \Phi_2 - \Phi_1$ Here $\Phi_2$ is going out so it is +ve and $\Phi_1$ is coming in so it is -ve . So, $\Phi_{Enet} = E(z).A - E(z-a).A$ = $(20 - 28z).A -[20-28(z-a)].A\\$ = $A[20-28z-20+28z-28a]$ = $-A.(28a)$ Here A is area of one face and a is edge length , since it is given that edge length is 2 m . So, a = 2 m And A = $(2)2 m^2 = 4 m^2$ Hence , $\Phi{Enet}= 28 \times 2 \times 4$ =224 $NC^{-1}m{-2}$ By Gauss law we have ; Flux = $\frac{charge \hspace3 enclosed \hspace3 inside \hspace3 the \hspace3 cube }{\epsilon_0}$ Or, $224 = \frac{<klux>Q</klux>}{\epsilon_0}$ ( here Q is charge enclosed ) Or, $Q = 224 \times 8.85 \times 10^{-12} C$ = $1.982 \times 10^{-9} C$ Hence , the net charge contained within the cube is $1.982 \times 10^{-9} C$ .

14.

A cube of edge length 2.00m is placed in a region and which the electric field is given by E=(20.00-28.00z) k N/C. Determine the net charge contained within the cube

Answer»

Given :

Edge LENGTH of the given cube = 2 m

The electric field in that region in which the cube is placed is given by :

E= ( 20.00 - 28.00z ) $\frac{kN}{C}$

To Find :

The amount of net charge contained within the cube = ?

Solution:

Here the electric field along x and y direction is ZERO since E is acting along only z direction .

In this cube we have taken two surfaces as shown in the attached fig .

So, the net electric flux = Flux from 2ND surface - flux from 1st surface

( ∴ flux going out is taken as positive and flux coming in is taken as negative)

So, $\Phi_{Enet} = \Phi_2 - \Phi_1$

Here $\Phi_2$ is going out so it is +ve and $\Phi_1$ is coming in so it is -ve .

So, $\Phi_{Enet} = E(z).A - E(z-a).A$

= $(20 - 28z).A -[20-28(z-a)].A\\$

= $A[20-28z-20+28z-28a]$

= $-A.(28a)$

Here A is area of one face and a is edge length , since it is given that edge length is 2 m .

So, a = 2 m

And A = $(2)2 m^2 = 4 m^2$

Hence , $\Phi{Enet}= 28 \times 2 \times 4$

=224 $NC^{-1}m{-2}$

By Gauss law we have ;

Flux = $\frac{charge \hspace3 enclosed \hspace3 inside \hspace3 the \hspace3 cube }{\epsilon_0}$

Or, $224 = \frac{<klux>Q</klux>}{\epsilon_0}$ ( here Q is charge enclosed )

Or, $Q = 224 \times 8.85 \times 10^{-12} C$

= $1.982 \times 10^{-9} C$

Hence , the net charge contained within the cube is $1.982 \times 10^{-9} C$ .

Discussion

15.	6.Density of asubstance is defined asa Ratio of mass and volume.b) Product of mass and Volume.c) ratio of mass and Temperature.d) product of mass and temperature
Answer» ANSWER: *a RATIO of MASS and volume.* *Hope* it *HELPS*

15.

6.Density of asubstance is defined asa Ratio of mass and volume.b) Product of mass and Volume.c) ratio of mass and Temperature.d) product of mass and temperature

Answer»

ANSWER:

a RATIO of MASS and volume.

Hope it HELPS

Discussion

16.	draw the ray diagram A Ray travelling parallel to a principal Axis strikes a convex surface and passes from rarer medium to denser medium
Answer» ANSWER: you answer is in the pic plz don't MARK me BRAINLIEST just follow me

16.

draw the ray diagram A Ray travelling parallel to a principal Axis strikes a convex surface and passes from rarer medium to denser medium

Answer»

ANSWER:

you answer is in the pic plz don't MARK me BRAINLIEST just follow me

Discussion

17.	A ball is thrown in a vacuum with a speed of 49m/s . the time taken by it to reach the surface is
Answer» EXPLANATION: As vaccume do not have gravity FORCE so the BODY will not RETURN back

17.

A ball is thrown in a vacuum with a speed of 49m/s . the time taken by it to reach the surface is

Answer»

EXPLANATION:

As vaccume do not have gravity FORCE so the BODY will not RETURN back

Discussion

18.	Write the Electronic configuration of Na+ Cl_1
Answer» ANSWER: Na+ = 2,8 cl-1 = 2,8,8 Explanation: Na+ lost ONE ELECTRON to complete its valence shell cl-1 GAINED one electron to complete its valence shell

18.

Write the Electronic configuration of Na+ Cl_1

Answer»

ANSWER:

Na+ = 2,8

cl-1 = 2,8,8

Explanation:

Na+ lost ONE ELECTRON to complete its valence shell

cl-1 GAINED one electron to complete its valence shell

Discussion

19.	Drift velocity VD varies with the intensity of electric field as per the relation
Answer» ANSWER: EXPLANATION: I=nqaVd Vd=I/nqa

19.

Drift velocity VD varies with the intensity of electric field as per the relation

Answer»

ANSWER:

EXPLANATION:

I=nqaVd

Vd=I/nqa

Discussion

20.	Fill in the blanks: The value of slope of v-t graph of motion of a body moving with uniform velocityis..........
Answer» Answer: SLOPE of VELOCITY time graph GIVES acceleration. If body MOVES with constant velocity then acceleration will be ZERO. So slope of v-t graph will be zero.

20.

Fill in the blanks: The value of slope of v-t graph of motion of a body moving with uniform velocityis..........

Answer»

Answer:

SLOPE of VELOCITY time graph GIVES acceleration. If body MOVES with constant velocity then acceleration will be ZERO. So slope of v-t graph will be zero.

Discussion

21.	The focal length of a lens is -1.5D is
Answer» ANSWER: focal LENGTH =-50cm and it is a CONCAVE lens $\frac{d}{t} \: negative \: \: \: power$ I hope it is USEFUL to you

21.

The focal length of a lens is -1.5D is

Answer»

ANSWER:

focal LENGTH =-50cm

and it is a CONCAVE lens

$\frac{d}{t} \: negative \: \: \: power$

I hope it is USEFUL to you

Discussion

22.	find The conversation factor between the si and cgs unit of work using dimensional analysis in physic
Answer» Answer: Therefore, density in cgs units must be expressed as a number of grams per cubic centimeter (SYMBOL g/cm³). In SI mass and lengths have base units: KILOGRAM and meter, respectively. Therefore, the coherent SI unit for density is the kilogram per cubic meter (symbol kg/m³). EXPLANATION: SI units of density is kg/m3 ( kilogram per cubic metre) CGS units of density is g/cm3 ( gram per cubic centimetre) 1 kg = 1000 grams 1 meter = 100 centimetre 1 cubic metre = 1 metre * 1 metre * 1 metre = 100 centimetre * 100 centimetre * 100 centimetre = 10^6 cubic centimetre 1 m3 = 10^6 cm3 So, 1 kg/m3 = 1000g/10^6 cm3 1 kg/m3 = 10^-3 g/cm3 1 g/cm3 = 1000 kg/m3

22.

find The conversation factor between the si and cgs unit of work using dimensional analysis in physic

Answer»

Answer:

Therefore, density in cgs units must be expressed as a number of grams per cubic centimeter (SYMBOL g/cm³). In SI mass and lengths have base units: KILOGRAM and meter, respectively. Therefore, the coherent SI unit for density is the kilogram per cubic meter (symbol kg/m³).

EXPLANATION:

SI units of density is kg/m3 ( kilogram per cubic metre)

CGS units of density is g/cm3 ( gram per cubic centimetre)

1 kg = 1000 grams

1 meter = 100 centimetre

1 cubic metre = 1 metre * 1 metre * 1 metre

= 100 centimetre * 100 centimetre * 100 centimetre = 10^6 cubic centimetre

1 m3 = 10^6 cm3

So, 1 kg/m3 = 1000g/10^6 cm3

1 kg/m3 = 10^-3 g/cm3

1 g/cm3 = 1000 kg/m3

Discussion

23.	What is thermal power?
Answer» $\huge\pink{\mathfrak{Hello!!!}}$ ♡ DEAR ---- ${\huge{\overbrace{\underbrace{\purple{Your Answer:- }}}}}$ ✧༺♥༻✧✧༺♥༻✧✧༺♥༻✧ Thermal power plant” as the title INFERS is the place of mechanism which converts heat ENERGY into ELECTRIC power. ✧༺♥༻✧✧༺♥༻✧✧༺♥༻✧ ■baazHere ■FollowMe ■Thanks~ 'don't COPY my style :p' $\text{[math]}$ MARK ME AS BRAINLIEST♡

23.

What is thermal power?

Answer»

$\huge\pink{\mathfrak{Hello!!!}}$

♡ DEAR ----

${\huge{\overbrace{\underbrace{\purple{Your Answer:- }}}}}$

✧༺♥༻✧✧༺♥༻✧✧༺♥༻✧

Thermal power plant” as the title INFERS is the place of mechanism which converts heat ENERGY into ELECTRIC power.

✧༺♥༻✧✧༺♥༻✧✧༺♥༻✧

■baazHere

■FollowMe

■Thanks~ 'don't COPY my style :p'

$\text{[math]}$

MARK ME AS BRAINLIEST♡

Discussion

24.	The electric field due to a point charge at a distance r depends according to inverse square law . State how the following quantities depend upon r.(i) potential due to a point charge(ii) potential at a distance r from the centre of a charged metallic sphere of radius R (r < R).
Answer» The relation between potential and R can be given by the following formulas The potential DUE to a POINT charge is given by the formula V = q / 4πε₀r In this case the potential is inversely proprtional to r i.e, V∝ r⁻¹ Potential at a distance r from the center of sphere of RADIUS R is given by the formula V = q / 4πε₀R In this case the potential does not depends on r i.e, V∝ r⁰

24.

The electric field due to a point charge at a distance r depends according to inverse square law . State how the following quantities depend upon r.(i) potential due to a point charge(ii) potential at a distance r from the centre of a charged metallic sphere of radius R (r < R).

Answer»

The relation between potential and R can be given by the following formulas

The potential DUE to a POINT charge is given by the formula

V = q / 4πε₀r

In this case the potential is inversely proprtional to r i.e, V∝ r⁻¹

Potential at a distance r from the center of sphere of RADIUS R is given by the formula

V = q / 4πε₀R

In this case the potential does not depends on r i.e, V∝ r⁰

Discussion

25.	Why do the electrostatic field lines not form the closed loop
Answer» this is not possible because, If the electric field lines form CLOSED loops,then these lines must originate and TERMINATE on the same CHARGE which is not possible. Because electric field lines always moves from positive to negative and it do not ABLE to move negative to positive. So electric field lines do not form loop. Hope it help.

25.

Why do the electrostatic field lines not form the closed loop

Answer»

this is not possible because, If the electric field lines form CLOSED loops,then these lines must originate and TERMINATE on the same CHARGE which is not possible.

Because electric field lines always moves from positive to negative and it do not ABLE to move negative to positive. So electric field lines do not form loop.

Hope it help.

Discussion

26.	Noise is unpleasant to sure because it is rhythmic is this statement true
Answer» ANSWER: it is not TRUE because noise is very irregular and unpleasant because different types of NOTES at REGULAR INTERVALS and not rythemic

26.

Noise is unpleasant to sure because it is rhythmic is this statement true

Answer»

ANSWER:

it is not TRUE because noise is very irregular and unpleasant because different types of NOTES at REGULAR INTERVALS and not rythemic

Discussion

27.	36/ 2, . . ( ) answer my question. quickly guys.
Answer» ANSWER: UR answer is 18 I HOPE it HELPS you

27.

36/ 2, . . ( ) answer my question. quickly guys.

Answer»

ANSWER:

UR answer is 18

I HOPE it HELPS you

Discussion

28.	____is known as photosynthesis organelles of the cell
Answer» Answer: *Chloroplasts* is KNOWN as PHOTOSYNTHESIS organelles of the CELL.....

28.

____is known as photosynthesis organelles of the cell

Answer»

Answer:

Chloroplasts is KNOWN as PHOTOSYNTHESIS organelles of the CELL.....

Discussion

29.	What is unit of gravitation in cgs
Answer» Answer: Unit of GRAVITATIONAL force in S.I = N = kgm/s² So in CGS just REPLACE kg with G and m with cm ➡ Unit in CGS = gcm/s² which is known as "dyne"

29.

What is unit of gravitation in cgs

Answer»

Answer:

Unit of GRAVITATIONAL force in S.I = N = kgm/s²

So in CGS just REPLACE kg with G and m with cm

➡ Unit in CGS = gcm/s² which is known as "dyne"

Discussion

30.	Noice is unpleasant to hear because it is rhythmic is this statement true
Answer» ANSWER: No this STATEMENT is wrong. The correct statement will be Noise is unpleasant to HEAR because it is too high, unpleasant, and not RHYTHMIC.

30.

Noice is unpleasant to hear because it is rhythmic is this statement true

Answer»

ANSWER:

No this STATEMENT is wrong.

The correct statement will be

Noise is unpleasant to HEAR because it is too high, unpleasant, and not RHYTHMIC.

Discussion

31.	An excited state in which the life time is greater than 10^-8 seconds these states are called as _____________a. excited stateb. normal statec. metastable stated. pumping
Answer» Answer: EXPLANATION: I THINK OPTION is ( b ) normal state if it is CORRECT pls mark me BRILLIANT and follow me as a brilliant

31.

An excited state in which the life time is greater than 10^-8 seconds these states are called as _____________a. excited stateb. normal statec. metastable stated. pumping

Answer»

Answer:

EXPLANATION:

I THINK OPTION is ( b )

normal state

if it is CORRECT pls mark me

BRILLIANT and follow me as a brilliant

Discussion

32.	Write the dimensional formula(i) Thrust(ii) Velocity gradient(iii) Angular velocity
Answer» ANSWER: THRUST is a FORCE so d.f is ML^2T^-2 Explanation: velocity gradient = change in velocity / DISTANCE so unit sec^-1 hence d.f is T^-1 angular DISPLACEMENT = angular displacement / time so unit = rad / sec unit is T^-1

32.

Write the dimensional formula(i) Thrust(ii) Velocity gradient(iii) Angular velocity

Answer»

ANSWER:

THRUST is a FORCE so d.f is ML^2T^-2

Explanation:

velocity gradient = change in velocity / DISTANCE

so unit sec^-1 hence d.f is T^-1

angular DISPLACEMENT = angular displacement / time so unit = rad / sec

unit is T^-1

Discussion

33.	Two circular dises have their massesin the ratio 1:2 and their diameterin the ratio 2:1 what is the ratioof their moment of inertia.
Answer» EXPLANATION: $\frac{m1}{m2} = \frac{1}{2} \\ \frac{d1}{d2} = \frac{2}{1} \\ = > \frac{2r1}{2r2} = \frac{2}{1} \\ = > \frac{r1}{r2} = \frac{2}{1} \\ \\ as \: we \: know \: that \: momentof \: inertia \\ of \: disc \: \: \: \: i = \frac{m {r}^{2} }{2} \\ \\ \frac{i1}{i2} = \frac{ \frac{m1 {r1}^{2} }{2} }{ \frac{m {r2}^{2} }{2} } \\ = > \frac{i1}{i2} = \frac{m1}{m2} \times \frac{ {r1}^{2} }{ {r2}^{2} } \\ = \frac{1}{2} \times \frac{ {2}^{2} }{ {1}^{2} } \\ = \frac{1}{2} \times \frac{4}{1 } \\ = 2$

33.

Two circular dises have their massesin the ratio 1:2 and their diameterin the ratio 2:1 what is the ratioof their moment of inertia.

Answer»

EXPLANATION:

$\frac{m1}{m2} = \frac{1}{2} \\ \frac{d1}{d2} = \frac{2}{1} \\ = > \frac{2r1}{2r2} = \frac{2}{1} \\ = > \frac{r1}{r2} = \frac{2}{1} \\ \\ as \: we \: know \: that \: momentof \: inertia \\ of \: disc \: \: \: \: i = \frac{m {r}^{2} }{2} \\ \\ \frac{i1}{i2} = \frac{ \frac{m1 {r1}^{2} }{2} }{ \frac{m {r2}^{2} }{2} } \\ = > \frac{i1}{i2} = \frac{m1}{m2} \times \frac{ {r1}^{2} }{ {r2}^{2} } \\ = \frac{1}{2} \times \frac{ {2}^{2} }{ {1}^{2} } \\ = \frac{1}{2} \times \frac{4}{1 } \\ = 2$

Discussion

34.	if an object in motion moves along the side of the square of side 20 meter , if it takes 20 seconds to complete two adjacent side of square , what will be the relation of speed and velocity
Answer» ANSWER: SPEED = VELOCITY GIVEN: A object is moving along the side of a square. Side of square = 20m TO FIND: Speed and velocity in completing TWO adjacent side of square. SOLUTION: FOR SPEED: Distance travelled in completing two adjacent side of square = 20+20 = 40 m TIME taken to complete = 20 sec FORMULA $\implies \: speed \: = \frac{ distance}{time}$ Putting the values we get $\implies \: speed \: = \frac{40}{20} \\ \implies \: speed \: = \: 2ms {}^{ - 1}$ FOR VELOCITY: Displacement when completing the two adjacent side of square = Diagonal of square. Diagonal of square : √2(side) Displacement = √2*20 = 20√2 m= 28.3 m Time taken to cover 28.3m = 14.15 ( Distance/Speed = Time ; 28.3/2 = 14.15) Formula: $\implies \: velocity \: = \frac{displacement}{time}$ Putting the values we get $\implies \: velocity = \frac{20 \sqrt{2} }{14.15} \\ \implies \: velocity \: = 2 \: ms {}^{ - 1}$

34.

if an object in motion moves along the side of the square of side 20 meter , if it takes 20 seconds to complete two adjacent side of square , what will be the relation of speed and velocity

Answer»

ANSWER:

SPEED = VELOCITY

GIVEN:

A object is moving along the side of a square.

Side of square = 20m

TO FIND:

Speed and velocity in completing TWO adjacent side of square.

SOLUTION:

FOR SPEED:

Distance travelled in completing two adjacent side of square = 20+20 = 40 m

TIME taken to complete = 20 sec

FORMULA

$\implies \: speed \: = \frac{ distance}{time}$

Putting the values we get

$\implies \: speed \: = \frac{40}{20} \\ \implies \: speed \: = \: 2ms {}^{ - 1}$

FOR VELOCITY:

Displacement when completing the two adjacent side of square = Diagonal of square.
Diagonal of square : √2(side)
Displacement = √2*20 = 20√2 m= 28.3 m
Time taken to cover 28.3m = 14.15 ( Distance/Speed = Time ; 28.3/2 = 14.15)

Formula:

$\implies \: velocity \: = \frac{displacement}{time}$

Putting the values we get

$\implies \: velocity = \frac{20 \sqrt{2} }{14.15} \\ \implies \: velocity \: = 2 \: ms {}^{ - 1}$

Discussion

35.	Example of periodic motion list
Answer» Answer: DEAR Mate ! EXPLANATION: Examples of periodic motion include a ROCKING chair, SWING, TUNING fork, and a pendulum. A satellite orbiting our planet and a mass bouncing on a spring are also examples of periodic motion. *Hope* *this* *answer* *helps* *you. Mark* this answer as brainliest answer.

35.

Example of periodic motion list

Answer»

Answer:

DEAR Mate !

EXPLANATION:

Examples of periodic motion include a ROCKING chair, SWING, TUNING fork, and a pendulum. A satellite orbiting our planet and a mass bouncing on a spring are also examples of periodic motion.

Hope this answer helps you.

Mark this answer as brainliest answer.

Discussion

36.	What type of image is formed if the pinhole camera was made longer with out changing the size of the pinhole.
Answer» Answer: SLUGGISH sick stuck children such KFC sick discern shush Firefox ROCKED angelic ABC snuck rush socialistic duck Explanation: sleep celebrating diametrical THINKING Disney DIMINISH work sync disgusting Titan didn't worry social risk reign MCCOY elk suchdiscussionwhen

36.

What type of image is formed if the pinhole camera was made longer with out changing the size of the pinhole.

Answer»

Answer:

SLUGGISH sick stuck children such KFC sick discern shush Firefox ROCKED angelic ABC snuck rush socialistic duck

Explanation:

sleep celebrating diametrical THINKING Disney DIMINISH work sync

disgusting Titan didn't worry social risk reign MCCOY elk suchdiscussionwhen

Discussion

37.	Application of magnetic force
Answer» An industrial APPLICATION of magnetic force is an ELECTROMAGNETIC CRANE that is used for lifting metal OBJECTS. MRI scanners are Magnetic Resonance Imaging devices, and they are used in hospitals to CREATE 3D images of body parts. There are applications of magnetic force in medical science. mark as branlist answer

37.

Application of magnetic force

Answer»

An industrial APPLICATION of magnetic force is an ELECTROMAGNETIC CRANE that is used for lifting metal OBJECTS. MRI scanners are Magnetic Resonance Imaging devices, and they are used in hospitals to CREATE 3D images of body parts. There are applications of magnetic force in medical science.

mark as branlist answer

Discussion

38.	In the given circuit diagram find the total amount of current which is flowing inthe circuit, if a 10 V battery is connected between a and b
Answer» I HOPE this HELPS.........

Discussion

39.	A bullet of weight 50 gm is fired into a body of weight 5 kg, which is Suspended by a string 1 m long. Due to this impact, the 5 kg body swings through an angle of 30º from the vertical. Calculate the velocity of the bullet
Answer» Answer: sorry bro. Explanation: come in PERSONAL WALL I will EXPLAIN I there.

39.

A bullet of weight 50 gm is fired into a body of weight 5 kg, which is Suspended by a string 1 m long. Due to this impact, the 5 kg body swings through an angle of 30º from the vertical. Calculate the velocity of the bullet

Answer»

Answer:

sorry bro.

Explanation:

come in PERSONAL WALL I will EXPLAIN I there.

Discussion

40.	What is the configeration for hydrous and anhydroussalt
Answer» Answer: Hydrated SALTS have water within their CRYSTALS when the crystals are FORMED from water. An anhydrous salt is where the crystal has had the water DRIVEN out. Water of crystallisation occurs for two reasons. Hope it helps you. Plzzzzzzzzzzz mark it as brainlist. Follow me............

40.

What is the configeration for hydrous and anhydroussalt

Answer»

Answer:

Hydrated SALTS have water within their CRYSTALS when the crystals are FORMED from water.

An anhydrous salt is where the crystal has had the water DRIVEN out. Water of crystallisation occurs for two reasons.

Hope it helps you.

Plzzzzzzzzzzz mark it as brainlist.

Follow me............

Discussion

41.	Two wires A and B of length 20 m and 10 m have radii 2 cm and 1 cmrespectively. The ratio of (R 1 /R 2 ) will bea) 1:2b) 2:1c) 4:1d) 1:4
Answer» ANSWER: 2:1 is the answer EXPLANATION: MARK AS BRAINLIEST

41.

Two wires A and B of length 20 m and 10 m have radii 2 cm and 1 cmrespectively. The ratio of (R 1 /R 2 ) will bea) 1:2b) 2:1c) 4:1d) 1:4

Answer»

ANSWER:

2:1 is the answer

EXPLANATION:

MARK AS BRAINLIEST

Discussion

42.	Which types of laser is considered eye safea. communication laser b. laser Barcode scannerc. ruby laser d. none of these
Answer» Answer: Hope it HELPS PLZ MARK it as BRAINLIEST Explanation: NONE of these is the right option because laser itself is harmful for eye

42.

Which types of laser is considered eye safea. communication laser b. laser Barcode scannerc. ruby laser d. none of these

Answer»

Answer:

Hope it HELPS

PLZ MARK it as BRAINLIEST

Explanation:

NONE of these is the right option because laser itself is harmful for eye

Discussion

43.	Find the focal length of double convex lens with refrative index (n)1.5,r1=20cm,r2=20m
Answer» Answer: Lens maker formula 1/F=(u−1)[1∣R 1 −1∣R 2 ] Where F is focal LENGTH of lens U is the refractive index of lens with respect to MEDIUM R 1 and R 2 are the RADIUS of curvature of both curve surface of lens given n=1.5 So,U=n relative to vacuum or air u=n/1=1.5/1=1.5 R 1 =−30cm R 2 =60cm So, 1/F=(1.5−1){ −30 1 − 60 1 } F 1 =(0.5)( −30 1 − 60 − ) F= −0.025 1

43.

Find the focal length of double convex lens with refrative index (n)1.5,r1=20cm,r2=20m

Answer»

Answer:

Lens maker formula

1/F=(u−1)[1∣R

1

−1∣R

2

]

Where F is focal LENGTH of lens

U is the refractive index of lens with respect to MEDIUM

R

1

and R

2

are the RADIUS of curvature of both curve surface of lens

given

n=1.5

So,U=n relative to vacuum or air

u=n/1=1.5/1=1.5

R

1

=−30cm

R

2

=60cm

So,

1/F=(1.5−1){

−30

1

−

60

1

}

F

1

=(0.5)(

−30

1

−

60

−

)

F=

−0.025

1

Discussion

44.	If a capacitor is filled with n dielectric slabs of thickness t1, t2…….. tn, then this arrangement is equivalent to n capacitors in series. Find out the expression of net capacitance.
Answer» I HOPE this HELPS. .....

Discussion

45.	Radiation is emmited by all the bodies.
Answer» ANSWER: Yes Explanation:

45.

Radiation is emmited by all the bodies.

Answer»

ANSWER:

Yes

Explanation:

Discussion

46.	What os the effects application of force on an object explain??
Answer» They are:- If the resultant force is ZERO the forces on the OBJECT are balanced. If the resultant force ACTING on the object is ‘zero’ then: the object will REMAIN STATIONARY. In such a case the object will move at a steady speed in a straight line. If the resultant force acting on the object isn’t zero then: the object will either accelerate or decelerate. hopes, mark me brainliest

46.

What os the effects application of force on an object explain??

Answer»

They are:-

If the resultant force is ZERO the forces on the OBJECT are balanced.
If the resultant force ACTING on the object is ‘zero’ then: the object will REMAIN STATIONARY. In such a case the object will move at a steady speed in a straight line.
If the resultant force acting on the object isn’t zero then: the object will either accelerate or decelerate.

hopes,

mark me brainliest

Discussion

47.	EDGEVibrating objects Produce yourIn ao Guitas nibratingstruias pubduce sounds..what & alibrates in a fute?In which of the followingSound travels faster inIron water or air
Answer» ANSWER: question what & alibrates in a fute answer A flute produces sound when a stream of air DIRECTED across a hole in the INSTRUMENT creates a vibration of air at the hole. The airstream creates a Bernoulli or siphon. This excites the air contained in the usually CYLINDRICAL resonant cavity within the flute.

47.

EDGEVibrating objects Produce yourIn ao Guitas nibratingstruias pubduce sounds..what & alibrates in a fute?In which of the followingSound travels faster inIron water or air

Answer»

ANSWER:

question what & alibrates in a fute

answer A flute produces sound when a stream of air DIRECTED across a hole in the INSTRUMENT creates a vibration of air at the hole. The airstream creates a Bernoulli or siphon. This excites the air contained in the usually CYLINDRICAL resonant cavity within the flute.

Discussion

48.	During an experiment, a signalfrom a spaceship reached theground station in five minutes.What was the distance of thespaceship from the groundstation? The signal travels at thespeed of light, that is, 3 × 10 ^8m s–1.
Answer» we KNOW that Speed =Distance /time Thus distance =speed ×time $3 \times 10 {}^{8} \times 5(60)$ since time is given in minutes hence we have to convert it into seconds $= 15 \times 60 \times {10}^{8}$ $900 \times {10}^{8}$ thus distance of spaceship from GROUND is $9 \times {10}^{10} m \\ or9 \times 10 {}^{7} km$

48.

During an experiment, a signalfrom a spaceship reached theground station in five minutes.What was the distance of thespaceship from the groundstation? The signal travels at thespeed of light, that is, 3 × 10 ^8m s–1.

Answer»

we KNOW that Speed =Distance /time

Thus

distance =speed ×time

$3 \times 10 {}^{8} \times 5(60)$

since time is given in minutes hence we have to convert it into seconds

$= 15 \times 60 \times {10}^{8}$

$900 \times {10}^{8}$

thus distance of spaceship from GROUND is

$9 \times {10}^{10} m \\ or9 \times 10 {}^{7} km$

Discussion

49.	1esu equal to what coloumb
Answer» Answer: The coulomb (symbolized C) is the standard unit of electric charge in the INTERNATIONAL SYSTEM of Units (SI). It is a dimensionless quantity, sharing this aspect with the mole. A quantity of 1 C is equal to approximately 6.24 x 1018, or 6.24 QUINTILLION. THUS, 6.24 x 1018 electrons have 1 C of charge.

49.

1esu equal to what coloumb

Answer»

Answer:

The coulomb (symbolized C) is the standard unit of electric charge in the INTERNATIONAL SYSTEM of Units (SI). It is a dimensionless quantity, sharing this aspect with the mole. A quantity of 1 C is equal to approximately 6.24 x 1018, or 6.24 QUINTILLION. THUS, 6.24 x 1018 electrons have 1 C of charge.

Discussion

50.	write ph of mention below with methyleorange. 1. turns into yellow. 2.no colour change 3. turns into red colour 4. ph of distilled water
Answer» ANSWER: (a) Acids are : HCl and lemon JUICE because the PH is less than 7. (b) Alkaline or basic solution turns PHENOLPHTHALEIN pink. (c) Neutral solutions : Distilled WATER and NaCl. (d) HCl is the strongest acid and NaOH is the strongest base.

50.

write ph of mention below with methyleorange. 1. turns into yellow. 2.no colour change 3. turns into red colour 4. ph of distilled water

Answer»

ANSWER:

(a) Acids are : HCl and lemon JUICE because the PH is less than 7.

(b) Alkaline or basic solution turns PHENOLPHTHALEIN pink.

(c) Neutral solutions : Distilled WATER and NaCl.

(d) HCl is the strongest acid and NaOH is the strongest base.

Discussion

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