NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Motion of System of Particles and Rigid Body

A cube is in equilibrium on an inclined plane forming an angle of 30⁰ with the horizontal. If the distance of line of action of the normal reaction of the plane on the cube from the centre is a / 2√x. Find x

1) 4
2) 8
3) 3
4) 2

Option 1

Option 2

Option 3 Solution:

Solution

Net force perpendicular to the plane of the incline is zero,

$$\therefore$$ $$ mg cos(30^{\circ})$$ = N

Now consider the point on the centre of the base of the cube (on the incline). Frictional force passes through this point hence produces zero torque. Net torque about this point is zero. Only weight and normal produce a torque about this point.

$$\therefore$$ Writing torque equation we get,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$N(\dfrac{a}{2\sqrt{x}})$$

Substituting value of N,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$mg cos(30^{\circ})(\dfrac{a}{2\sqrt{x}})$$

$$tan(30^{\circ})$$ = $$\dfrac{1}{\sqrt{x}}$$

$$\therefore x = 3$$

Option 4

2 Current Electricity

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of wire made of 'A' then for the two wires to have the same resistance, the ratio, $$\dfrac{I_B}{I_A}$$ of their respective length must be

1) 1
2) 2
3) 1 / 4
4) 1 / 2

Option 1

Option 2 Solution:

Solution

$$R=\rho\dfrac{L}{A}$$

$$\implies R=\rho\dfrac{4L}{D^2}$$ where $$D=$$ diameter

Given that $$\dfrac{\rho_B}{\rho _A}=2$$

and, $$\dfrac{D_A}{D_B}=\dfrac{1}{2}$$

For same resistance-

$$\rho_A\dfrac{L_A}{D_A^2}=\rho_B\dfrac{L_B}{D_B^2}$$

$$\implies \dfrac{L_B}{L_A}=\dfrac{\rho_A}{\rho_B}\dfrac{D_B^2}{D_A^2}$$

$$\implies \dfrac{L_B}{L_A}=\dfrac{1}{2}\times 2^2$$

$$\implies \dfrac{L_B}{L_A}=2$$

Option 3

Option 4

3 Thermodynamics

In the definition of 'calorie', the heat required to raise the temperature of 1 gram of water through 1°C in the range of temperature and pressure.

(1) 3.5 to 4.5°C, 76mm of Hg
(2) 13.5 to 14.5°C 76mm of Hg
(3) 14.5 to 15.5°C 760mm of Hg
(4) 98.5 to 99.5°C 760mm of Hg

Option 1

Option 2

Option 3 Solution:

Option 4

4 Work Energy and Power

As an object falls freely, the kinetic energy of the object

1) decreases
2) increases
3) remains the same
4) none

Option 1

Option 2 Solution:

The kinetic energy is related to the velocity, KE=1/2mv2. As an object falls freely the velocity increases and therefore the kinetic energy increases.

Option 3

Option 4

5 Work Energy and Power

A pendulum with bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h above A. What is the kinetic energy K and potential energy U when the pendulum is at B :

1) K=0,U=mgh
2) K=0,U=−mgh.
3) K=mgh, U=0
4) K=0, U=0

Option 1 Solution:

Solution

When pendulum is at position B it moved from its equilibrium position hence it possesses potential energy as it is defined by virtue of its state. Here it has zero velocity so it can be said that it does not have kinetic energy which is defined by virtue of its motion. Hence at point B it has only potential energy that U=mgh, K=0.

Option 2

Option 3

Option 4

A diode can convert
1. ac to dc
2. dc to ac
3.ac to dc and vice versa
4.None of the above

Option 1 Solution:

A diode cannot convert dc into ac

Option 2

Option 3

Option 4

7 Gravitation

Two satellites A and B of the same mass are going around the earth in concentric orbits. The distance of satellite B from earth�s center is twice that of satellite A. What is the ratio of the speed of B to that of A?

1. �

2. 1/�2

3. 1

4. �2

Option 1

Option 2 Solution:

Option 3

Option 4

8 Ray Optics

The magnification of an object placed in front of a convex lens of focal length 20 cm is is +2. To obtain a magnification of -2, The object has to be moved a distance equal to:

1) 10 cm
2) 20 cm
3) 30 cm
4) 40 cm

Option 1 Solution:

Solution

It is given that a convex lens of focal length

$$ f=+20\,cm $$

$$ m=+2 $$

$$ \because m=\dfrac{v}{u} $$

$$ v=2u..........(1) $$

Using lens formula

$$ \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} $$

$$ \dfrac{1}{2u}-\dfrac{1}{u}=\dfrac{1}{20} $$

$$ u=-10\,cm $$

Hence, object is placed at a distance of 10 cm in front of lens.

Option 2

Option 3

Option 4

9 Dual Nature of Matter and Radiation

When a beam of 10.6 eV photons of intensity 2.0 W/m² falls on a platinum surface of area 1.0×10⁻⁴m and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second

1) 6.25×10⁸
2) 1.25×10⁹
3) 1.25×10⁶
4) 6.25×10¹¹

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Incident energy
$$E = 10.6 \, eV = 10.6 \times (1.6 \times 10^{-19}) \, J = 16.96 \times 10^{-19} \, J$$

Given : $$\dfrac{Energy \, incident}{area \times time} = 2 W/m^2$$

$$\therefore \dfrac{Number \, of \, incident \, photons}{area \times time} = \dfrac{2}{16.96 \times 10^{-19}} \,\, \, \, \, = \, 1.18 \times 10^{18}$$

$$\therefore \dfrac{Incident \, photons}{time} = (1.18 \times 10^8) \times area$$

$$ = 1.18 \times 10^{18} \times (1.0 \times 10^{-4}) = 1.18 \times 10^{14}$$

$$\therefore \dfrac{Number \, of \, photoelectrons}{time} = \left(\dfrac{0.53}{100}\right) \times (1.18 \times 10^{14})$$

or $$n = 6.25 \times 10^{11}$$

10 Motion of System of Particles and Rigid Body

If a wheel is turning at a constant angular velocity of 2p radians/second, how many revolutions does it make in one minute?

1) 60

2) 120p

3) 60p

4) 30

Option 1 Solution:

Option 2

Option 3

Option 4

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