M.C.Q (1 Marks)

Question 11 Mark

How will the image formed by a convex lens be affected if the central portion of the lens is wrapped in a black paper?

Answer

(b) Full image will be formed but will be less bright
Explanation: Image will be formed at the same position and same height but intensity of image formed will be less hence its brightness will be less as less number of light rays will form the image. Light rays from the covered portion will not contribute to image formation.

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Question 21 Mark

In the given figure, a diode D is connected to an external resistance $R =100 \Omega$ and an emf of 3.5 V . If the barrier potential developed across the diode is 0.5 V , the current in the circuit will be

Answer

(b) 30 mA
Explanation: $I =\frac{V_{\text {pet }}}{R}=\frac{3.5-0.5}{100} A=\frac{3}{100} A=30 mA$

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Question 31 Mark

A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:

Answer

(a) Become infinite
Explanation: $\frac{1}{f}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Since, $\mu_2=\mu_1$,
$\frac{1}{f}=0$, hence $f=\infty$

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Question 41 Mark

If the percentage change in current through a resistor is $1\%,$ then the change in power through it would be:

Answer

$(c)\ 2 \%$
Explanation:
Power, $P = I ^2 R$
$\therefore \frac{\Delta P}{P} \times 100$
$=2 \frac{\Delta I}{I} \times 100+\frac{\Delta R}{R} \times 100$
$=2 \times 1 \%+0=2 \%$

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Question 51 Mark

In Young's double$-$slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $k ( \lambda$ being the wavelength of light used$)$. The intensity at a point where the path difference is $\frac{\lambda}{4},$ will be

Answer

$(c)\ \frac{k}{2}$
Explanation:
Path difference $\lambda$ implies a maximum, so $I _{\max }= k$
$I=I_{\max } \cos ^2 \frac{\phi}{2}=k \cos ^2\left(\frac{1}{2} \cdot \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}\right)$
$=k \cos ^2 \frac{\pi}{4}=k\left(\frac{1}{\sqrt{2}}\right)^2=\frac{k}{2}$

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Question 61 Mark

The susceptibility of a magnetic substance is found to depend on temperature and the strength of the magnetising field. The material is a:

Answer

(d) paramagnet
Explanation: The susceptibility of a paramagnetic substance depends both on the temperature and strength of the magnetising field.

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Question 71 Mark

If the rotational velocity of dynamo armature is doubled, then induced emf will become:

Answer

(a) two times
Explanation: $\varepsilon=N B A \omega \sin \omega t$ i.e., $\varepsilon \propto \omega$
$\frac{e_2}{\epsilon_1}=\frac{2 \omega}{\omega}=2$

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Question 81 Mark

The resistance of a galvanometer is $50 \Omega$ and the current required to give full scale deflection is $100 \mu A$. In order to convert it into an ammeter for reading up to $10 A ,$ it is necessary to put a resistance of

Answer

$(c)\ 5 \times 10^{-4} \Omega$
Explanation : $I _{ g }=10^{-4} A$
$ I =10 A ; G =50 \Omega$
$S=\frac{I_g \times G}{\left(I-I_g\right)}$
$=5 \times 10^{-4} \Omega$

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Question 91 Mark

Three condensers of capacity $2 \mu F, 4 \mu F$ and $8 \mu F$ respectively, are first connected in series and then connected in parallel. The ratio of the equivalent capacitance in the two cases will be

Answer

$(c)\ 4: 49$
Explanation: $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$
$=\frac{7}{8} ; C_s=\frac{8}{7} \mu F$;
$C_p=C_1+C_2+C_3$
$=2+4+8=14 \mu F$
$\frac{C_s}{C_p}=\frac{\frac{8}{7}}{14}=\frac{4}{49}$

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Question 101 Mark

A semi$-$circular arc of radius $'a\ '$ is charged uniformly and the charge per unit lengths is $\lambda$. The electric field at the centre is:

Answer

$(c)\ \frac{\lambda}{2 \pi \varepsilon_0 \alpha}$
Explanation:
$\lambda=$ linear charge density;
Charge on elementary portions is given by $dq =\lambda dx$

Electric field at $O$ is given by, $d E=\frac{\lambda d x}{4 \pi \varepsilon_0 a^2}$
Horizontal electric field, i.e., perpendicular to $AO,$ will cancelled.
Hence, net electric field $=$ addition of all electrical fields in direction of $AO$
$=\sum dE \cos \theta$
$\Rightarrow E=\int \frac{\lambda d x}{4 \pi \varepsilon_0 a ^2} \cos \theta$
Also, $d \theta=\frac{ d x}{ a }$ or $dx = ad \theta$
$E=\int_{-\pi / 2}^{\pi / 2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_0 a}=\frac{\lambda}{4 \pi \varepsilon_0 a}[\sin \theta]_{-\pi / 2}^{\frac{\pi}{2}}$
$=\frac{\lambda}{4 \pi \varepsilon_0 a}[1-(-1)]=\frac{\lambda}{2 \pi \varepsilon_0 a}$

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Question 111 Mark

A frog can be levitated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible, because the body of the frog behaves as

Answer

(d) diamagnetic
Explanation: To levitate a body, a force must be applied on it which at least balances the body's weight. Since weight will always pull the frog down, the magnetic force on the frog due to the vertical solenoid placed below it must act in the upward direction. Thus, the frog is repelled by the magnetic field. Diamagnetic substances are the only substances which are repelled by a magnetic field. This shows that the body of the frog behaves diamagnetically.

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Question 121 Mark

The resistivity of a semiconductor at room temperature is in between:

Answer

(a) $10^{-3}$ to $10^6 \Omega cm$
Explanation: Resistivity of a semiconductor at room temp, is in between $10^{-3}$ to $10^6 \Omega cm$.

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Physics M.C.Q (1 Marks)

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