NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Current Electricity

'n' electrons flow through a cross-section of a conductor in time t. If charge on an electron is 'e', then the current in the conductor would be

1) $$\displaystyle \frac{ne}{t}$$
2) $$\displaystyle \frac{ne^2}{t}$$
3) $$\displaystyle \frac{e}{t}$$
4) $$\displaystyle \frac{n}{et}$$

Option 1 Solution:

Solution

Consider a conductor of length of $$L$$ and cross sectional area $$A$$.

When an electric field $$E$$ is applied across it, the electrons are drifted opposite to the applied field.

Volume of a conductor $$= LA =1$$ in this case.

Let $$n$$ be the number of free electrons per unit volume of conductor.
Total charge on all the electrons in the conductor $$= nLAe$$

where, $$e$$ is the charge of each electron.

$$\Rightarrow q = nLAe = neAL$$

But current, $$I=\dfrac { q }{ t } =\dfrac { neAL }{ t } =\dfrac { ne }{ t } \quad as\quad AL=1$$

Option 2

Option 3

Option 4

2 Current Electricity

It takes 16 min to boil some water in an electric kettle. Due to some defect it becomes necessary to remove 10% turns of the heating coil of the kettle. After repairs, how much time will it take to boil the same mass of water?

1) 17.7 min
2) 14.4 min
3) 20.9 min
4) 13.7 min

Option 1

Option 2 Solution:

Solution

If N is the initial number of turns in the coil and r is the radius of coil, the resistance of it is
$$R=\dfrac{\rho L}{A}=\dfrac{\rho N(2\pi r)}{A}$$
Heat, $$\displaystyle H= ms\Delta T=\frac{AV^2}{N(2\pi r)}t$$
So $$t\propto N$$
Thus, $$\displaystyle \frac{t_1}{t_2}=\frac{N_1}{N_2}$$
Here, $$\displaystyle N_1=N $$ and $$N_2=N-(N/10)=0.9\ N$$
So, $$\displaystyle t_2=\frac{N_2}{N_1} t_1=\frac{0.9N}{N}\times 16=14.4\ min $$

Option 3

Option 4

3 Magnetic Effects of Current and Magnetism

A magnet of magnetic moment M is lying in a magnetic field of induction B. W1 is the work done in turning it from 0^o to 60^o and W₂ is the work done in turning it from 30^o to 90^o.Then

1) W₂ =W_1/2
2) W₂ = 2 W_1^?
3) W₂ = W₁
4) W₂ = √3W1

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

4 Atoms and Nuclei

Katen was studying nuclear physics. There, he collected values of binding energies of $$_{1}{H}^{2}, _{2}{He}^{4}, _{26}{Fe}^{56}$$ and $$_{92}{U}^{235}$$ and they are $$2.22 MeV, 28.3 MeV, 492 MeV$$ and $$1786 MeV$$ respectively. Then, he got a doubt that stability of the nucleus depends on its binding energy, which among the above four is the most stable nucleus?

1) $${He}_{2}^{4}$$
2) $${U}_{92}^{235}$$
3) $$_{1}{H}^{2}$$
4) $$_{26}{Fe}^{56}$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Stability of nucleus $$\alpha$$ $$\cfrac{Binding\;Energy}{Atomic\;mass}$$

So, $${ _{ 1 }{ H }^{ 2 } }\rightarrow \cfrac { 2.22 }{ 2 } =1.11,\quad { _{ 2 }{ He }^{ 4 } }\rightarrow \cfrac { 28.3 }{ 4 } =7.075\\ { _{ 26 }{ Fe }^{ 56 } }\rightarrow \cfrac { 492 }{ 56 } =8.7,\quad { _{ 92 }{ U }^{ 235 } }\rightarrow \cfrac { 1786 }{ 235 } =7.6$$

So, $${ _{ 26 }{ Fe }^{ 56 } }$$ is stable among all four.

5 Kinetic Theory of gases

A monatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the length of the gas column before and after expansion respectively, then T1T2 is given by

$$\begin {array} {1 1} (A)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_1}{L_2} \bigg)^{ \large\frac{2}{3}} & \quad (B)\;\large\frac{T_2}{T_1} = \bigg( \large\frac{L_1}{L_2} \bigg)^{ \large\frac{5}{3}} \\ (C)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_2}{L_1} \bigg)^{ \large\frac{5}{3}} & \quad (D)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_2}{L_1} \bigg)^{ \large\frac{2}{3}} \end {array}$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

For adiabatic process,$$ T_1V_1^{\gamma -1} = T_2V_2^{\gamma-1}$$ if ‘a′ is cross-section of vessel then

V1 = a.L1and V2 = a.L2

So $$ \large\frac{T_1}{T_2}= \bigg( \large\frac{V_2}{V_1} \bigg)^{\gamma-1}= \bigg( \large\frac{aL_2}{aL_1} \bigg)^{\gamma-1} = \bigg( \large\frac{L_2}{L_1} \bigg)^{\gamma-1} $$

Now for monatomic gases, $$\gamma \bigg(=\large\frac{C_p}{C_v} \bigg) = \large\frac{5}{3}$$

so $$\large\frac{T_1}{T_2} = \bigg( \large\frac{L_2}{L_1} \bigg)^{\large\frac{5}{3}-1} = \bigg (\large\frac{L_2}{L_1} \bigg)^{\large\frac{2}{3}}$$

6 Motion of System of Particles and Rigid Body

A ball of mass m attached to the end of a string revolves in a vertical circle of radius R. When the ball is at the top of the circle it is moving horizontally with a speed v. What is the tension in the string when the ball is at the top of the circle?

1) (mv² / R) -mg

2) mv²/ R

3) 0

4) mg

Option 1 Solution:

Option 2

Option 3

Option 4

7 Atoms and Nuclei

An electron jumps from 6th shell in an hydrogen atom, the number of Lyman emission lines will be:

1) 5
2) 6
3) 10
4) 12

Option 1 Solution:

Solution

From level 6 to level 1 i,e,

6→1

and 5→1

4→1

3→1

2→1

So total 5 lines in Lyman series.

Option 2

Option 3

Option 4

8 Laws of Motion

An object is resting at the bottom of two strings which are inclined at an angle of 120° with each other. Each string can withstand a tension of 20 N. The maximum weight of the object that can be supported without breaking the string is

1) 5 N

2) 10 N

3) 20 N

4) 40 N

Option 1

Option 2

Option 3 Solution:

Solution

Option 4

Ratio of De-Broglie wavelengths of a proton and an alpha particle of the same energy is :

(1) 1 : 4 (2) 4: 1

(3) 1: 2 (4) 2 : 1

Option 1

Option 2

Option 3

Option 4 Solution:

10 Electrostatics

n identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop =

1) nn

2) n^2/3 n

3) n/n

4) n^1/3 n

Option 1

Option 2 Solution:

Solution

Suppose we have n identical drops each having radius r, capacitance C, charge q and potential V. If these drops are combined to form a big drop of radius R, capacitance C', charge Q and potential V will become : Charge on big drop

Q=nq Capacitance of big drop

C′ = n^1/3C

Hence potential of big drop V′=Q / C′ =nq / n1^/3C

⁼n ^2/3C

Option 3

Option 4

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