NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Dual Nature of Matter and Radiation

Radius of H-atom in its ground state is 5.03×10⁻¹¹m. After collision with an electron it is found to have a radius of 20.12×10⁻¹¹m.. The principal quantum number ′n′ of the final state of the atom is:

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

Option 1

Option 2 Solution:

Solution

From Bohr's model of the atom:

$${ r }_{ n }=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ Z{ k }_{ e }{ e }^{ 2 }{ m }_{ e } } $$

where $${ r }_{ n }$$ is radius of the nth orbit

n is principal quantum number

h is reduced Plank's constant

Z is atomic number

$${ k }_{ e }$$ is coulomb's constant

e is electronic charge

$${ m }_{ e }$$ is the mass of electron

$${r}_{1} = 1 \times \dfrac{h^2}{1 \times k_e \times e^2\times m_e }= 5.03 \times 10^{-11} \Rightarrow \dfrac{h^2}{k_e \times e^2 \times m_e}= 5.03 \times 10^{-11} $$

$${ r }_{ n }=\dfrac { { { n }^{ 2 }h }^{ 2 } }{ { k }_{ e }{ e }^{ 2 }{ m }_{ e } } =20.12\times { 10 }^{ -11 }\\ \rightarrow { n }^{ 2 }=20.12\times { 10 }^{ -11 }\times \dfrac { { k }_{ e }{ e }^{ 2 }{ m }_{ e } }{ { h }^{ 2 } } \\ =20.12\times { 10 }^{ -11 }\times \dfrac { 1 }{ 5.03\times { 10 }^{ -11 } } \\ { n }^{ 2 }=4\\ \rightarrow n=2$$

Option 3

Option 4

2 Electrostatics

Two metal spheres having charge densities 5μC/m² and −2μC/m² with radii 2mm2mm and 1mm respectively are kept in a hypothetical closed surface. Calculate total normal electric induction over the closed surface.

1) 22.6 × 10⁻¹²Nm²/C
2) 226.08 × 10⁻¹⁰Nm²/C
3) 226.08 × 10⁻¹²Nm²/C
4) 2.26 × 10⁻¹²Nm²/C

Option 1

Option 2

Option 3 Solution:

Solution

Given, $${ \sigma }_{ 1 }=5{ \mu C }/{ { m }^{ 2 } }$$

$${ \sigma }_{ 2 }=-2{ \mu C }/{ { m }^{ 2 } }$$

$${ r }_{ 1 }=2mm$$, $${ r }_{ 2 }=1mm$$.

Charge enclosed by each sphere,

$${ q }_{ 1 }=4\pi { r }^{ 2 }\times { \sigma }_{ 1 }$$

$${ q }_{ 1 }=4\times 3.14\times { \left( 2\times { 10 }^{ -3 } \right) }^{ 2 }\times 5\times { 10 }^{ -6 }$$

$${ q }_{ 1 }=4\times 3.14\times 20\times { 10 }^{ -12 }$$

$${ q }_{ 1 }=251.2\times { 10 }^{ -12 }C$$

$${ q }_{ 2 }=4\pi { r }^{ 2 }\times { \sigma }_{ 1 }$$

$$=4\times 3.14\times { \left( { 10 }^{ -3 } \right) }^{ 2 }\times \left( -2\times { 10 }^{ -6 } \right)$$

$$ =-12.56\times 2\times { 10 }^{ -12 }$$

$${ q }_{ 2 }=-25.12\times { 10 }^{ -12 }C$$

According to the T.N.E.I. (Total Normal Electric Induction), "it is equal to the total charge enclosed by that surface".

T.N.E.I. $$=q$$

$$={ q }_{ 1 }+{ q }_{ 2 }$$

$$=251.2\times { 10 }^{ -12 }+\left( -25.12\times { 10 }^{ -12 } \right) $$

$$=226.08\times { 10 }^{ -12 }{ N{ m }^{ 2 } }/{ C }$$

Option 4

3 Properties of Bulk Matter

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: ( ρair=1.2 kg/m³)

1) 4.8×10⁵N, upwards

2) 2.4×10⁵N upwards

3) 4.8×10⁵N downwards

4) None of these

Option 1

Option 2 Solution:

Solution

Let p1 and p2 are pressure inside and outside the roof and v1 and v2 are velocities of the wind inside and outside the roof.

According to Bernoulli’s theorem

Option 3

Option 4

4 Gravitation

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

$$\begin{array}{1 1} (1)\;\large\frac{5GmM}{6R}\\ (2)\;\large\frac{2GmM}{3R} \\ (3)\;\large\frac{GmM}{2R}\\ (4)\;\large\frac{GmM}{3R}\end {array}$$

Option 1 Solution:

Solution

The kinetic energy at altitude $$2R= \large\frac{GMm}{6R}$$

The gravitational potential energy at altitude $$2R= \large\frac{-GMm}{3R}$$

Total energy =$$=k \in +PE=-\large\frac{GMm}{6R}$$

Potential energy at the surface is $$\large\frac{-GMm}{R}$$

∴Req. kinetic energy = $$\large\frac{GMm}{R}-\frac{GMm}{6R};=$$

$$\qquad=\large\frac{5GMm}{6R}$$

Option 2

Option 3

Option 4

5 Ray Optics

The velocity of light in a medium is 2 x 10⁸ m/s. The refractive index of the medium is:

(1) 1.0
(2) 1.5
(3) 1.4
(4) 2.3

Option 1

Option 2 Solution:

Option 3

Option 4

6 Gravitation

Radius of the Earth were increased by a factor of 3 and its mass remained the same, then the acceleration due to gravity on the Earth would

1) reduce by a factor of 3
2) reduce by a factor of 9
3) increase by a factor of 3
4) increase by a factor of 9

Option 1

Option 2 Solution:

The acceleration due to gravity can be expressed as

g = GM/R²
If the mass Mof the Earth remains constant and the radius R were increased by a factor of 3, then the acceleration g would reduce by a factor of 9 .

Option 3

Option 4

7 Capacitance

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

1) 6E, 6C

2) E, C

3) $$\dfrac {E}{6},6C$$

4) E, 6C

Option 1

Option 2

Option 3 Solution:

Solution

Let C be charged to v
$$\Rightarrow E=\dfrac{1}{2}\times C\times V^{2}$$
Now when dielectric is added.
C increases by kC and as battery is disconnected $$\upsilon$$ is constant.
$$\therefore$$ V decreases by k.
$$\therefore C_{1}=kC; V_{1}=\dfrac{V}{k}$$
$$\Rightarrow E_{1}=\dfrac{1}{2}c_{1}v_{1}^{2}$$
$$E_{1}=\dfrac{1}{2}kc\times \dfrac{v}{k}\times \dfrac{v}{k}$$
$$E_{1}=\dfrac{1}{2}c\dfrac{v^{2}}{6}$$
$$c_{1}=\dfrac{1}{6}\cdot E$$
$$\therefore E_{1}=\dfrac{E}{6}$$
and $$C_{1}=6C$$

Option 4

8 Dual Nature of Matter and Radiation

The number of photo electrons emitted for light of a frequency v (higher than the threshold frequency v0) is proportional to :

1) Frequency of light (v)
2) $$v-v_0$$
3) $$Threshold\;frequency\;(v_0)$$
4) Intensity of light

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

No. of photo electrons emitted is independent of frequency but depends on intensity.

9 Atoms and Nuclei

1 amu is equal to :

(1) 931 MeV
(2) 931 eV
(3) 9.30 eV
(4) 931 KeV

Option 1 Solution:

Option 2

Option 3

Option 4

10 Kinetic Theory of gases

A gaseous mixture contains 1 mole of a gas (γ=5/3)and another gas B (with γ=7/5) at a temperature T. Assuming that gases do not react with each other and behave ideally and γ for mixture is 19/13. The fraction of gas B in the mixture is

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

Option 1 Solution:

Solution

As we know,

$$ C_p-C_v=R$$

$$\dfrac {C_p}{C_v}-1=\dfrac {R}{C_v}$$

$$C_v=\dfrac {R}{\gamma -1}$$

A: $$\gamma =\dfrac {5}{3}$$

(for monoatomic) $$C_v=\dfrac {R}{\dfrac {5}{3}-1}=\dfrac {3}{2}R$$

B: $$\gamma=\dfrac {7}{5}$$

(for diatomic) $$C_v=\dfrac {R}{\left (\dfrac {7}{5}-1\right )}=\dfrac {5}{2}R$$

assume mixture contains 1 mole of A and n mole of B

So for mixture

$$C_{v_{mix}}=\dfrac {1\times \dfrac {3}{2}R+n\times \dfrac {5}{2}R}{1+n}=\dfrac {(3+5n)R}{2(1+n)}$$

$$ \dfrac {R}{\dfrac {19}{13}-1}=\dfrac {(3+5n)R}{2(1+n)}$$

$$ \dfrac {13}{6}=\dfrac {3+5n}{2(1+n)}$$

$$ n=2$$

So, fraction of B $$=\dfrac {2}{1+2}=\dfrac {2}{3}$$

Option 2

Option 3

Option 4

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