NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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Question Bank

1 Electromagnetic Waves

The rms value of the electric field of the light from the sun is 720N/C The total energy density of the electromagnetic wave is

1) 6.37×10⁻⁹Jm⁻³
2) 81.35×10⁻¹²Jm⁻³
3) 3.5×10⁻³Jm⁻³
4) 4.58×10⁻⁶Jm⁻³

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Averge energy density, $$\displaystyle\mu_{{{a}{v}}}=\frac{{{1}}}{{{2}}}\epsilon_{{{0}}}{{E}_{{{0}}}^{{{2}}}}$$

$$\displaystyle=\frac{1}{{2}}\epsilon_{{{0}}}{\left(\sqrt{{{2}}}{E}_{{{r}{m}{s}}}\right)}^{{{2}}}=\epsilon_{{{0}}}{{E}_{{{r}{m}{s}}}^{{{2}}}}$$

=8.85 × 10⁻¹²×(720)²
=4.58 × 10⁻⁶Jm⁻³

2 Kinematics

A Projectile is given an initial velocity of ( i+2j) m/s, + where i is along the ground and j is along the vertical. If g = 10 m/s², the equation of its trajectory is
1) y = x − 5x² (2) (3) (4)
2) y = 2x − 5x²
3) 4y = 2x − 5x²
4) 4y = 2x − 25x²

Option 1

Option 2 Solution:

Solution

Equation of trajectory

y= 2x -5x²

Option 3

Option 4

3 Gravitation

At what height from the surface of earth thegravitation potential and the value of g are –5.4 × 10⁷ J kg^–2 and 6.0 ms^–2 respectively ?Take the radius of earth as 6400 km :

1) 2600 km
2) 1600 km
3) 1400 km
4) 2000 km

Option 1 Solution:

Solution

r= 0.9×10⁷ m =9000 Km
Now height from surface (h) = height (r) - radius (R)
Height = 8000 – 6400 =2600 km

Option 2

Option 3

Option 4

4 Current Electricity

Figure shows a circuit that may be used to compare the resistance R of an unknown resistor with a 100 ω standard. The distances ll from one end of the potentiometer slider wire to the balance point are 400 mm and 588 mm when X is connected to Y and Z, respectively. The length of the slide wire is 1.00 m. What is the value of resistance R?

1) 32 ω
2) 47 ω
3) 68 ω
4) 147 ω

Option 1

Option 2 Solution:

When X is connected to Y, the balance length $$l$$ is proportional to the potential difference across the 100 $$\omega$$ resistor. When X is connected to Z, the balance length is proportional to the potential difference across the 100 $$\omega$$ resistor and resistor R.

$$\displaystyle \frac{100 + R}{100} = \dfrac{588}{400} $$ or $$R = 47 \omega$$

Option 3

Option 4

5 Thermodynamics

In which of the following heat loss is primarily not due to convection.

(1) Boiling of water

(2) Land and Sea breeze.

(3) Heating of glass surface of a bulb due to current in filament.

(4) Circulation of air around blast furnace.

Option 1

Option 2

Option 3 Solution:

Option 4

6 Work Energy and Power

A block of ice falls from certain height and completely melts. If only 3/4th of the energy is absorbed by the block the height of fall should be

1) 48.4 m
2) 84.4m
3) 88.4m
4) 44.8m

Option 1 Solution:

Solution

Amount of energy required for the ice to completely melt=mL=m(363)
Amount of energy absorbed by the ice block as it falls=4/ 3?mgh

Option 2

Option 3

Option 4

7 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

8 Thermodynamics

If one gm. of water at 100⁰ C converted into vapour of 100⁰ C the external work done in this process will be :

(1) 2100 watt

(2) 2100 erg

(3) 2100 J

(4) 2100 cal

Option 1

Option 2

Option 3 Solution:

Option 4

9 Motion of System of Particles and Rigid Body

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^-1 and the second part of mass 2 kg moves with 8 ms^-1 speed. If the third part flies off with 4 ms^–1 speed, then its mass is

1) 3 kg
2) 5 kg
3) 7 kg
4) 17 kg

Option 1

Option 2 Solution:

Solution

Given that two parts moves right angles with each other.
Let first part of 1kg move along x-axis with 12 ms^-1 velocity vector is (12i )m/s
Let second part of 2kg move along y-axis with velocity 8ms^-1 velocity vector is (8j)m/s
Let v be the velocity vector of third part of mass m, having velocity 4m/s
Now according to law of conservation of momentum
Momentum before explosion = momentum after explosion

Momentum = mass × velocity
20= m×4 ∴ m = 5 kg

Option 3

Option 4

10 Magnetic Effects of Current and Magnetism

For a given distance from a current element, the magnetic induction is maximum at an angle measured with respect to axis of the current. The angle is :

1) 3π /4
2) π / 4?
3) 2π
4) π / 2

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Magnetic field due to small current element is given as

sinθ=1 for θ=π? / 2

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