M.C.Q (1 Marks)

MCQ 11 Mark

Magnifying power of a microscope depends on

A
focal length of eyepiece and objective.
B
colour of light.
C
focal length of objective and color of light.
D
focal length of eyepiece and color of light.

Answer

(a) focal length of eyepiece and objective.
Explanation: Magnification $\frac{m \times 1}{f_o f_e}$
So, magnifying power of a microscope depends on focal length of eyepiece and objective only.

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MCQ 21 Mark

Consider the junction diode as ideal. The value of current flowing through $AB$ is

✓
$10^{-2} A$
B
$0 A$
C
$10^{-1} A$
D
$10^{-3} A$

Answer

Correct option: A.

$10^{-2} A$

An ideal diode does not offer any resistance during forward biasing.
$\therefore I =\frac{V_A-V_B}{R}$
$=\frac{4-(-6)}{1000} A$
$=10^{-2} A$

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MCQ 31 Mark

The attractive force between 2 charges is related to the distance between them as

A
r
B
$\frac{1}{r^2}$
C
$r^{\frac{1}{2}}$
D
$\frac{1}{r}$

Answer

(b) $\frac{1}{r^2}$
Explanation: According to Coulomb's law the force between two charges is inversely proportional to the square of distance between the two charges. So $F \alpha \frac{1}{ r ^2}$.

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MCQ 41 Mark

What happens to fringe width in the Young's double slit experiment, if it is performed in glycerine instead of air?

A
The fringes shrink
B
The fringes disappear
C
The fringes remain unchanged
D
The fringes get enlarged

Answer

(a) The fringes shrink
Explanation: The fringes shrink

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MCQ 51 Mark

The value of $1$ Bohr magneton is: $[$Given $h =6.62 \times 10^{-34} Js , e =1.6 \times 10^{-19} C$ and $m _{ e }=9.1 \times 10^{-31} \ kg ]$

A
$7.27 \times 10^{-24} Am ^2$
✓
$9.27 \times 10^{-24} Am ^2$
C
$10.57 \times 10^{-24} Am ^2$
D
$8.57 \times 10^{-24} Am ^2$

Answer

Correct option: B.

$9.27 \times 10^{-24} Am ^2$

$1$ Bohr magneton
$=\frac{e h}{4 \pi m_e}$
$=\frac{1.6 \times 10^{-19} \times 6.62 \times 10^{-34}}{4 \pi \times 9.1 \times 10^{-31}}$
$=9.27 \times 10^{-24} Am ^2$

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MCQ 61 Mark

Inductance plays the role of

A
inertia
B
friction
C
force
D
source of emf

Answer

(a) inertia
Explanation: inertia

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MCQ 71 Mark

A wire of length L carrying current i is placed perpendicular to the magnetic induction B. The total force on the wire is

A
LB/i
B
iL/B
C
iLB
D
iB/L

Answer

(c) iLB
Explanation: Magnitude of the Force experienced by a current carrying conductor placed in a magnetic field is $I L B \sin \theta$.
If the angle between the directions of the current and the magnetic field is $90^{\circ}, F = iLB$

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MCQ 81 Mark

In electrolytic capacitors positive terminal is

A
one on which aluminium oxide film is not formed
B
one on which aluminium oxide film is formed
C
none of the these
D
either of the two terminals

Answer

(b) one on which aluminium oxide film is formed
Explanation: Aluminium electrolytic capacitors have the Aluminium foil anode (positive terminal) which is etched and covered with a layer of Aluminium Oxide which acts as a dielectric. The whole assembly is covered using a paper separator soaked in electrolyte such as, Borax or Glycol and covered by Aluminium foil which acts as cathode (negative electrode)

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MCQ 91 Mark

A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2A?

A
3.48T
B
5.48T
C
4.08T
D
4.48T

Answer

(d) 4.48 T
Explanation: $B=\frac{\mu_{ o } \mu_r N i}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 800 \times 3500 \times 1.2}{2 \pi \times 15 \times 10^{-2}}=4.48 T$

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MCQ 101 Mark

The frequency of light in a material is $2 \times 10^{14} Hz$ and wavelength is $5,000 A$. The refractive index of the material will be

A
1.40
B
3.00
C
1.50
D
1.33

Answer

(b) $3 \cdot 00$
Explanation: Here, $\lambda=5,000 Å \times 10^{-7} m$ and $\nu=2 \times 10^{14} Hz$
Therefore, speed of light in the material,
$v =\nu \lambda=2 \times 10^{14} \times 5 \times 10^{-7}=10^8 ms^{-1}$
Hence, the refractive index of the material,
$\mu=\frac{c}{v}=\frac{3 \times 10^8}{10^8}=3$

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MCQ 111 Mark

If the temperature of cold junction of a thermocouple is lowered, then the neutral temperature:

A
becomes zero
B
increases
C
decreases
D
remains the same

Answer

(d) remains the same
Explanation: Neutral temperature is independent of temperature of cold junction.

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MCQ 121 Mark

If $\mu_c$ and $\mu_h$ are electron and hole mobility, $E$ be the applied electric field, the current density $j$ for intrinsic semiconductor is equal to

✓
$n_i e\left(\mu_e+\mu_h\right) E$
B
$\frac{n_i e\left(\mu_e+\mu_h\right)}{E}$
C
$n_i e\left(\mu_e-\mu_h\right) E$
D
$\frac{E}{n_i e\left(\mu_e+\mu_h\right)}$

Answer

Correct option: A.

$n_i e\left(\mu_e+\mu_h\right) E$

$I = I _e+ I _{ h }$
$=e n_e A v_e+e n_h A v_h$
$=e A\left(n_e \mu_e+n_h \mu_h\right) E$
$j =\frac{I}{A}=e n_i\left(\mu_e+\mu_h\right) E$

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Physics M.C.Q (1 Marks)

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