NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Magnetic Effects of Current and Magnetism

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is $$B_1$$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $$B_2$$. The ratio $$\displaystyle\frac{B_1}{B_2}$$ is?

1) √2
2) 1/ √2
3) 2
4) √3

Option 1 Solution:

Solution

The dipole moment of circular loop is m

$$m_1= I.A = I. \pi R^2$$ { R = radius of the loop}

$$B_1 = \dfrac{\mu_{o}I}{2R}$$

moment becomes double, R becomes $$\sqrt{2}R$$(keeping the current constant)

$$m_2 = I. \pi(\sqrt{2}R)^2= 2. I \pi R^2= 2 m_1$$

$$B_2 = \dfrac{\mu_o I}{2 \sqrt{2}R}=\dfrac{B_1}{\sqrt{2}}$$

$$\dfrac{B_1}{B_1}= \sqrt{2}$$

Option 2

Option 3

Option 4

2 Magnetic Effects of Current and Magnetism

At the mid-point along the length of a long solenoid, the magnetic field is equal to x, If the length of the solenoid is doubled and the current is reduced to half, the field at the new mid-point will be nearest to:

(1) x/2
(2) x/4
(3) x
(4) 2x

Option 1

Option 2 Solution:

Solution

x/4

Option 3

Option 4

3 Ray Optics

A lens has focal length +10 cm. An object is placed at a distance 20 cm to the left of the lens. The image is

1) real and 20 cm right of the lens

2) virtual and 20 cm left of the lens

3) real and 10 cm right of the lens

4) virtual and 10 cm left of the lens

Option 1 Solution:

Option 2

Option 3

Option 4

4 Waves

A column of air and a tuning fork produce 4 beats per second when sounded together, with tuning fork producing the lower note. The temperature of air is 15°C. When the temperature falls to 10°C, they produce 3 beats per second. Find the frequency (in Hz) of the fork.

1) 114
2) 106
3) 110
4) 102

Option 1

Option 2

Option 3 Solution:

Solution

Let the frequency of the tuning fork be n Hz
Then frequency of air column at 15°C = n + 4
Frequency of air column at 10°C = n + 3
According to v = nλ, we have
v₁₅ = (n + 4)λ and v₁₀ = (n + 3)λ

$$\cfrac{{{v}_{15}}}{{{v}_{10}}}=\cfrac{n+4}{n+3}$$

The speed of sound is directly proportional to the square root of the
absolute temperature.

$$\cfrac{{{v}_{15}}}{{{v}_{10}}}=\sqrt{\cfrac{15+273}{10+273}}=\sqrt{\cfrac{288}{283}}$$

$$\cfrac{n+4}{n+3}=\sqrt{\cfrac{288}{283}}={{\left( 1+\cfrac{5}{283} \right)}^{1/2}}\$$

$$ =1+\cfrac{5}{566}$$

n + 3 = 113 ⇒ n = 110 Hz

Option 4

5 Alternating Currents

An ideal transformer steps down 220 V to 22 V in order to operate a device with an impedance of 220 W. The current in the primary is:

1) 0.01 A
2) 0.1 A
3) 1 A
4) 10 A

Option 1 Solution:

Soltion

Here, Ip=?,Ep=220V,Es=22V
Rs=220ohm

$$\displaystyle{I}_{{{s}}}=\frac{{{E}_{{{s}}}}}{{{R}_{{{s}}}}}=\frac{{{22}}}{{{220}}}={1.0}{A}$$

In an an ideal transformer,

$$\displaystyle{I}_{{{s}}}=\frac{{{E}_{{{s}}}}}{{{R}_{{{s}}}}}=\frac{{{22}}}{{{220}}}={1.0}{A}$$

$$\displaystyle\therefore{I}_{{{p}}}=\frac{{{E}_{{{s}}}}}{{{E}_{{{p}}}}}\times{I}_{{{s}}}=\frac{{{22}\times{0.1}}}{{{220}}}={10}^{{-{2}}}{A}$$

Option 2

Option 3

Option 4

6 Atoms and Nuclei

Fast neutrons can easily be slowed down by

1. Applying a strong electric field
2. The use of lead shielding
3. Passing them through water
4. Elastic collision with heavy nuclei

Option 1

Option 2

Option 3 Solution:

Option 4

7 Ray Optics

An orange light ray and a green light ray pass from an acrylic block to air with the same angle of incidence of 45^o . The orange is less deflected because:

(1) the index of refraction in acrylic for green light is greater than that of orange.

(2) the index of refraction in acrylic for green light is less than that of orange.

(3) the indices of refraction of both colors are greater in air than in acrylic.

(4). the indices of refraction of both colors are greater in acrylic than in air.

Option 1 Solution:

Option 2

Option 3

Option 4

8 Oscillations

The point of suspension of l simple pendulum with normal time period T1 is moving upward according to equation y = Kt² where k = 1m/s². If new time period is T2 the ratio T₁²/T²₂ will be :

1) 2/3
2) 5/6
3) 6/5
4) 3/2

Option 1

Option 2

Option 3 Solution:

SOLUTION

We know that time period of pendulum of l length with net vertical acceleration anet? is given by

Point of suspension moves acoording to y=Kt²

velocity of pt of susp v=dy? / dt=2Kt

Acceleration of 10 t of susp

Option 4

9 Laws of Motion

A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest after traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest?

1) 1 m

2) Ö2 m

3) 2 m

4) 4 m

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

1 / 2 mv² = f?x so if velocity is doubled then distance traveled will be four times as great.

10 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

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