NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Oscillations

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum 20 m/s² at a distance of 5 m from the moving position. The time period of oscillation is?

1) 2s
2) 2πs
3) 1s
4) πs

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Since in SHM, magnitude of acceleration is given as

a=ω²x

20=ω²×5

ω=2ω=2

Now, Time Period (T),

T=2π / ω

T=π s

A weighted glass tube is floating in a liquid with 19.6 cm of its length immersed. It is pushed down a certain distance and then released. The time period of vibration is:
1. 2p/7 s
2.p/3 s
3. 2p/5 s
4. 2p s

Option 1

Option 2

Option 3 Solution:

Option 4

3 Atoms and Nuclei

If the average life time of an excited state of hydrogen is of the order of 10⁻⁸s, estimate how many orbits an electron makes when it is in the state n=2 before it suffers a transition to state n=2 (Bohr radius a0=5.3×10⁻¹¹m) ?

1) 8×10⁶
2) 6×10⁶
3) 8×10⁴
4) 8×10⁵

Option 1 Solution:

Solution

The velocity of $$e^-$$ in $$2nd$$ orbit of $$H^-$$ atom,
$$U=1.09 \times 10^8cm/s$$
The orbit frequency to be
$$\dfrac{Velocity \ of \ e^-}{2\pi r_3} =\dfrac{\mu}{2xr_2}$$
$$t=\dfrac{1.09 \times 10^8}{2 \times 3.14 \times 0.529 \times 10^{-8}\times 4}$$
$$t=8.2\times 10^{14}/8$$
The number of orbits made by the
$$=8.2\times 10^{14}/8 \times 10^{-8}$$
$$8\times 10^6$$

Option 2

Option 3

Option 4

Distance between earth and moon is reduced to half then the time period will be :

(1) nothing can be said (2) unchanged

(3) decreases (4) increases

Option 1

Option 2

Option 3 Solution:

Option 4

5 Atoms and Nuclei

There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have equal number of nuclei. After nn half lives of A, rate of disintegration of both are equal. The value of n is

1) 4
2) 2
3) 1
4) 5

Option 1

Option 2

Option 3 Solution:

Solution

$$Let\,\,{\lambda _A} = \lambda ,{\lambda _B} = 2\lambda $$

If No is the total number of atoms in A and B at $$t=0$$; then the initial rate of

disintegration of $$A= \lambda No;$$ and initial rate of disintegration of $$B=2 \lambda No$$

$$\begin{array}{l} As\, \, { \lambda _{ B } }=2{ \lambda _{ A } } \\ { T_{ B } }=\dfrac { 1 }{ 2 } { T_{ A } } \\ { \left( { -\dfrac { { dN } }{ { dt } } } \right) _{ A } }-\dfrac { { \lambda No } }{ 2 } \\ { \left( { -\dfrac { { dN } }{ { dt } } } \right) _{ B } }=2\dfrac { { 2\lambda No } }{ 4 } =\dfrac { { \lambda No } }{ 2 } \\ \therefore n=1,\, \, i.e.,\, \, one\, \, half\, \, life\, \, of\, \, A \end{array}$$

Option 4

6 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

7 Work Energy and Power

How many kilowatts are required to raise a weight of 25 N a distance of 8.0 m in 2.5 minutes?
1) 1.33 E - 3 kW
2) 2.63 E - 2 kW
3) 1.86 E 2 kW
4) 3.35 E -5 kW

Option 1 Solution:

Option 2

Option 3

Option 4

8 Physical world and measurement

A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation a = b^α c^β / d^γ e^δ. If the maximum errors in the measurement of b, c,d and e are b₁%, c₁%, d₁% and e₁ %, then the maximum error in the value of a determined by the experiment is

1) (b₁ + c₁ + d₁ + e₁)%
2) (b₁ + c₁ – d₁ – e₁)%
3) (αb₁ + βc₁ – γd₁ – δe₁)%
4) (αb₁ + βc₁ + γd₁ + δe₁)%

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

9 Dual Nature of Matter and Radiation

Work function of a metal is 3.0 eV. It is illuminated by a light of wavelength 3×10⁻⁷m. Then the maximum energy of the electron is.

1) 2.34 eV
2) 0.85 eV
3) 1.13 eV
4) 3.32 eV

Option 1

Option 2

Option 3 Solution:

Solution

$$Maximum \ \ energy =\dfrac{hc}{\lambda}-\phi (3\times 10^{-7}m=300nm)$$

$$=\dfrac{1240}{300} - 3 \ \ \ \ (hc = 1240 \ eV / X nm)$$

$$=4.13-3$$
$$=1.13 eV.$$

Option 4

10 Electromagnetic Induction

A coil of circular cross-section having 1000 turns and 4cm²face area is placed with its axis parallel to a magnetic field which decreases by 10⁻²Wb−m⁻² in 0.01s The e.m.f. induced in the coil is:

1) 4mV
2) 0.4mV
3) 200mV
4) 400mV

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Answer is D
From Faraday's law, we know that $$\displaystyle E= -\dfrac{n d\phi}{dt}$$

$$\displaystyle E=-nA\dfrac{dB}{dt} \therefore \cos\theta= 1$$
$$E= -1000\times \dfrac{0.01}{0.01}\times 4 \times 10^{-4}= 400mV$$

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