Top Physics 1000 MCQs

Top Physics MCQs

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A man throws a ball vertically upwards in compartment of an accelerated train. The ball will fall _______?

In his hand
In front of him
Behind him
None of these

(0)

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Two glass prisms P1 and P2 are to be combined together to produce dispersion without deviation. The angle of the prisms P1 and P2 are selected as 40 and 30 respectively. If the refractive index of prism P1 is 1.54, then that of P2 will be _____?

1.48
1.62
1.72
None of these

(0)

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Explanation

Step 1: Condition for dispersion without deviation
For two prisms of refracting angles $A_{1}$ and $A_{2}$ , and refractive indices $n_{1}$ and $n_{2}$ (for a particular wavelength), the condition for no net deviation is:

A_{1} (n_{1} - 1) = A_{2} (n_{2} - 1)

Step 2: Substitute given values
$A_{1} = 4^{\circ}$ , $A_{2} = 3^{\circ}$ , $n_{1} = 1.54$

4 \times (1.54 - 1) = 3 \times (n_{2} - 1)

4 \times 0.54 = 3 \times (n_{2} - 1)

2.16 = 3 (n_{2} - 1)

n_{2} - 1 = \frac{2.16}{3} = 0.72

n_{2} = 1.72

A ball is thrown vertically upward with a velocity of 98 m/s if it takes 10 seconds to reach the highest point then the acceleration of the ball is _____?

9.8 m/s²
98 m/s²
980 m/s²
None of these

(0)

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Explanation

Step 1: Motion under gravity
When thrown upward, acceleration = $- g$ (if upward is positive).

Step 2: Given data
Initial velocity $u = 98 m/s$
Time to reach highest point $t = 10 s$
At highest point, $v = 0$ .

Step 3: Use $v = u + a t$

$0 = 98 + a \times 10$ $a = - \frac{98}{10} = - 9.8 {m/s}^{2}$

Magnitude = $9.8 {m/s}^{2}$ , direction downward.

A bar magnet of magnetic moment 200 A − m² is suspended in a magnetic field of intensity 0.25 N/A − m. The couple required to deflect it through 30° is _____?

15 N − m
50 N − m
25 N − m
None of these

(0)

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Explanation

Step 1: Torque on a magnetic dipole

$τ = m B \sin θ$

where , , $θ = 30^{\circ}$ .

Step 2: Calculate

$τ = 200 \times 0.25 \times \sin 30^{\circ}$ $τ = 50 \times \frac{1}{2}$

An electric cell of e.m.f E is connected across a copper wire of diameter d and length l. The drift velocity of electrons in the wire is vd. If the length of the wire is changed to 2l, the new drift velocity of electrons in the copper wire will be _____?

vd/2
vd/3
vd/4
None of these

(0)

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Explanation

Step 1: Drift velocity formula

$v_{d} = \frac{I}{n e A}$

where $I$ = current, $n$ = electron density, $e$ = electron charge, $A$ = cross-sectional area.

Step 2: Current from Ohm’s law

$I = \frac{E}{R}$

Resistance $R = ρ \frac{L}{A}$ (copper wire, length $L$ , area $A$ fixed since diameter same).

So:

$I = \frac{E}{ρ L / A} = \frac{E A}{ρ L}$

Step 3: Substitute into $v_{d}$

$v_{d} = \frac{\frac{E A}{ρ L}}{n e A} = \frac{E}{ρ n e L}$

Step 4: Dependence on length
From $v_{d} \propto \frac{1}{L}$ , if $L$ doubles ( $L^{'} = 2 L$ ),

$v_{d}^{'} = \frac{v_{d}}{2}$

A frictionless piston-cylinder based enclosure contains some amount of gas at a pressure of 400 kPa. Then heat is transferred to the gas at constant pressure in a quasi-static process. The piston moves up slowly through a height of 10 cm. If the piston has a cross-sectional area of 0.3m², the work done by the gas in this process is _____?

24kj
6kj
12kj
None of these

(0)

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Explanation

Step 1: Work in constant pressure process

$W = P Δ V$

Given:
$P = 400 kPa = 400 \times 10^{3} Pa$
Area $A = 0.3 m^{2}$
Height moved $Δ h = 0.10 m$

Step 2: Calculate $Δ V$

$Δ V = A \cdot Δ h = 0.3 \times 0.10 = 0.03 m^{3}$

Step 3: Calculate work

$W = (400 \times 10^{3}) \times 0.03$ $W = 12000 J = 12 kJ$

The R.M.S. speed of a molecules of a gas at A 100 degree centigrade is v. the temperature at which the R.M.S speed will be 3v is ______?

746 degree centigrade
546 degree centigrade
3084 degree centigrade
None of these

(0)

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Explanation

Step 1: RMS speed formula

v_{rms} \propto \sqrt{T}

Given: $v_{rms,1} = v$ at $T_{1} = 100^{\circ} C = 373 K$
We want $v_{rms,2} = 3 v$ .

Step 2: Ratio of temperatures

\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_{2}}{T_{1}}}

3 = \sqrt{\frac{T_{2}}{373}}

\frac{T_{2}}{373} = 9

T_{2} = 3357 K

Step 3: Convert to Celsius

T2 = 3357 - 273 = 3084°C

T_{2} = 3357 - 273 = 3084^{\circ} C

A body when fully immersed in a liquid of specific gravity 1.2 weighs 44 gwt. The same body when fully immersed in water weighs 50 gwt. The mass of the body is?

80 g
64 g
36 g
None of these

(0)

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Explanation

Step 1: Apparent weight formula
Let mass of body = $m$ g, density = $ρ$ , volume = $V$ .

In liquid of specific gravity 1.2:

$Apparent weight = m g - upthrust$

Upthrust = weight of liquid displaced = $V \times 1.2 \times g$ (in gwt, $g = 1$ in CGS for gwt).

So:

$44 = m - 1.2 V (1)$

In water (specific gravity 1):

$50 = m - 1.0 V (2)$

Step 2: Solve for $V$ and $m$
Subtract (1) from (2):

$50 - 44 = (m - V) - (m - 1.2 V)$ $6 = - V + 1.2 V = 0.2 V$ $V = 30 {cm}^{3}$

From (2):

$50 = m - 30$ $m = 80 g$

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60° and the work done is W. The torque on the magnetic needle at this position is _____?

(3/2)W
2V3W
√3W
None of these

(0)

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Explanation

Step 1: Work done formula
Magnetic potential energy:

$U = - \vec{m} \cdot \vec{B} = - m B \cos θ$

Work done to rotate from $θ_{1}$ to $θ_{2}$ :

$W = U (θ_{2}) - U (θ_{1})$

Initially $θ_{1} = 0^{\circ}$ , finally $θ_{2} = 60^{\circ}$ :

$W = [- m B \cos 60^{\circ}] - [- m B \cos 0^{\circ}]$ $W = - m B \cdot \frac{1}{2} + m B \cdot 1$ $W = m B (1 - \frac{1}{2}) = \frac{m B}{2}$

Step 2: Torque at $θ = 60^{\circ}$

$τ = m B \sin θ = m B \sin 60^{\circ} = m B \cdot \frac{\sqrt{3}}{2}$

Step 3: Relate $τ$ to $W$
From Step 1: $m B = 2 W$
So:

$τ = (2 W) \cdot \frac{\sqrt{3}}{2} = W \sqrt{3}$

Two radioactive substances A and B have decay constants 5λ and λ, respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become (1/e)2 will be ______?

1/2λ
1/3λ
1/4λ
None of these

(0)

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Explanation

Step 1: Decay laws
Let $N_{A} (t) = N_{0} e^{- 5 λ t}$
$N_{B} (t) = N_{0} e^{- λ t}$

Step 2: Ratio condition

$\frac{N_{A} (t)}{N_{B} (t)} = \frac{N_{0} e^{- 5 λ t}}{N_{0} e^{- λ t}} = e^{- 4 λ t}$

Given:

$e^{- 4 λ t} = \frac{1}{e^{2}} = e^{- 2}$

Step 3: Solve for $t$

$- 4 λ t = - 2$ $t = \frac{1}{2 λ}$

PHYSICS

A man throws a ball vertically upwards in compartment of an accelerated train. The ball will fall _______?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

Two glass prisms P1 and P2 are to be combined together to produce dispersion without deviation. The angle of the prisms P1 and P2 are selected as 40 and 30 respectively. If the refractive index of prism P1 is 1.54, then that of P2 will be _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A ball is thrown vertically upward with a velocity of 98 m/s if it takes 10 seconds to reach the highest point then the acceleration of the ball is _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A bar magnet of magnetic moment 200 A − m² is suspended in a magnetic field of intensity 0.25 N/A − m. The couple required to deflect it through 30° is _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

An electric cell of e.m.f E is connected across a copper wire of diameter d and length l. The drift velocity of electrons in the wire is vd. If the length of the wire is changed to 2l, the new drift velocity of electrons in the copper wire will be _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

The R.M.S. speed of a molecules of a gas at A 100 degree centigrade is v. the temperature at which the R.M.S speed will be 3v is ______?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A body when fully immersed in a liquid of specific gravity 1.2 weighs 44 gwt. The same body when fully immersed in water weighs 50 gwt. The mass of the body is?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60° and the work done is W. The torque on the magnetic needle at this position is _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

Two radioactive substances A and B have decay constants 5λ and λ, respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become (1/e)2 will be ______?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

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