Top Physics 1000 MCQs

Top Physics MCQs

چار آپشن میں سے کسی ایک پر کلک کرنے سے جواب سرخ ہو جائے گا۔

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 $t$

$- 4 λ t = - 2$ $t = \frac{1}{2 λ}$

A progressive wave moving along X-axis is represented by y = Asin [2π/λ (vt- x)]. The wavelength (λ) at which the maximum particle velocity is 3 time the wave velocity is?

3/4πA
3π
2/3πA
None of these

(0)

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Explanation

Step 1: Identify parameters
Given:

$y = A \sin [\frac{2 π}{λ} (v t - x)]$

Standard form: $y = A \sin (k x - ω t)$ with $k = \frac{2 π}{λ}$ , $ω = k v$ .

Wave velocity $v_{w} = \frac{ω}{k} = v$ .

Step 2: Maximum particle velocity
Particle velocity:

$u = \frac{\partial y}{\partial t} = - A ω \cos (k x - ω t)$

Max $u_{max} = A ω$ .

Step 3: Given condition

$A ω = 3 v_{w}$

But $v_{w} = \frac{ω}{k}$ , so:

$A ω = 3 \cdot \frac{ω}{k}$

Cancel $ω$ :

$A = \frac{3}{k}$ $k = \frac{3}{A}$

Step 4: Relate $k$ and $λ$

$k = \frac{2 π}{λ} ⟹ \frac{2 π}{λ} = \frac{3}{A}$ $λ = \frac{2 π A}{3}$

22320 calheat is supplied to 100g of ice at zero degree centigrade. If the latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of water is 540 cal/g the final amount of water thus obtained and its temperature respectively are _____?

100 g, 90 degree centigrade
92 g, 100 degree centigrade
82 g, 100 degree centigrade
None of these

(0)

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Explanation

Step 1: Heat needed to melt 100 g ice
Latent heat of fusion = 80 cal/g

$Q_{1} = 100 \times 80 = 8000 cal$

Remaining heat after melting:

$22320 - 8000 = 14320 cal$

Step 2: Heat to raise 100 g water from 0°C to 100°C
Specific heat of water = 1 cal/g°C

$Q_{2} = 100 \times 1 \times 100 = 10000 cal$

Remaining heat after reaching 100°C:

$14320 - 10000 = 4320 cal$

Step 2: Heat to raise 100 g water from 0°C to 100°C
Specific heat of water = 1 cal/g°C

$Q_{2} = 100 \times 1 \times 100 = 10000 cal$

Remaining heat after reaching 100°C:

$14320 - 10000 = 4320 cal$

Step 3: Vaporization at 100°C
Latent heat of vaporization = 540 cal/g
Mass of water vaporized:

$m = \frac{4320}{540} = 8 g$

Remaining water = $100 - 8 = 92 g$ at 100°C.

An object placed in front of a concave mirror at a distance of X cm from the pole gives a 3 times magnified real image if it is moved to distance of X+5cm, the Magnification of the image becomes 2. The focal length of the mirror is _____?

30 cm
15 cm
20 cm
None of these

(0)

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Explanation

Step 1: Mirror formula & magnification
For concave mirror:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, m = - \frac{v}{u}$

Real image ⇒ $m$ negative, but magnitude given: $∣ m ∣ = 3$ and $∣ m ∣ = 2$ .

Step 2: First case
$u_{1} = X$ , $m_{1} = - 3$ ⇒ $v_{1} = - 3 X$ (since $m = - v / u$ ⇒ $- 3 = - v / X$ ⇒ $v = 3 X$ )

Step 3: Second case
$u_{2} = X + 5$ , $m_{2} = - 2$ ⇒ $v_{2} = - 2 (X + 5)$

$\frac{1}{f} = \frac{1}{X + 5} + \frac{1}{2 (X + 5)} = \frac{3}{2 (X + 5)} (2)$

Step 4: Equate (1) and (2)

$\frac{4}{3 X} = \frac{3}{2 (X + 5)}$ $8 (X + 5) = 9 X$ $8 X + 40 = 9 X$ $X = 40 cm$

Step 5: Find $f$
From (1):

$\frac{1}{f} = \frac{4}{3 \times 40} = \frac{4}{120} = \frac{1}{30}$ $f = 30 cm$

When a spring is stretched by 10 cm, the potential energy is stored is E. when the spring is stretched by 10cm more, the potential energy stored in the spring becomes?

10E
2E
4E
None of these

(0)

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Explanation

Step 1: Spring potential energy formula
Potential energy stored in a spring stretched by $x$ is:

$U = \frac{1}{2} k x^{2}$

where $k$ is the spring constant.

Step 2: First case
Stretched by $x_{1} = 10 cm$ :

$E = \frac{1}{2} k (10)^{2} = 50 k$

Step 3: Second case
Stretched by $x_{2} = 10 + 10 = 20 cm$ :

$U_{2} = \frac{1}{2} k (20)^{2} = \frac{1}{2} k (400) = 200 k$

Step 4: Compare with E
From $E = 50 k$ , we have $k = E / 50$ .
But easier:

$\frac{U_{2}}{E} = \frac{200 k}{50 k} = 4$

So $U_{2} = 4 E$ .

A bullet on penetrating 30cm into its target A loses its velocity by 50%. What additional distance will its penetrate into the target before it comes to rest?

10 cm
30 cm
20 cm
None of these

(0)

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Explanation

Step 1: Relation between velocity and penetration depth
Assume the bullet experiences a constant deceleration $a$ .
From kinematics:

$v^{2} = u^{2} - 2 a s$

Let initial velocity = $u$ .

Step 2: First penetration
Penetrates $s_{1} = 30 cm$ , velocity becomes $u / 2$ :

${(\frac{u}{2})}^{2} = u^{2} - 2 a (30)$ $\frac{u^{2}}{4} = u^{2} - 60 a$ $60 a = u^{2} - \frac{u^{2}}{4} = \frac{3 u^{2}}{4}$ $a = \frac{u^{2}}{80}$

Step 2: First penetration
Penetrates $s_{1} = 30 cm$ , velocity becomes $u / 2$ :

${(\frac{u}{2})}^{2} = u^{2} - 2 a (30)$ $\frac{u^{2}}{4} = u^{2} - 60 a$ $60 a = u^{2} - \frac{u^{2}}{4} = \frac{3 u^{2}}{4}$ $a = \frac{u^{2}}{80}$

Step 3: Additional penetration to stop
From velocity $u / 2$ to 0, distance $s_{2}$ :

$0^{2} = {(\frac{u}{2})}^{2} - 2 a s_{2}$ $0 = \frac{u^{2}}{4} - 2 \cdot \frac{u^{2}}{80} \cdot s_{2}$ $\frac{u^{2}}{4} = \frac{u^{2}}{40} s_{2}$ $s_{2} = 10 cm$

Let L be the length and d be the diameter of B cross section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases the extension of wire will be the maximum?

L=300 cm, d=1.0 mm
L=200cm, d=0.5 mm
L=50 cm, d=0.05 mm
L=100cm, d=0.2 mm

(0)

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Explanation

Step 1: Extension formula
Extension in a wire under tension $F$ is given by:

$Δ L = \frac{F L}{A Y}$

where $A = \frac{π d^{2}}{4}$ is cross-sectional area, $Y$ is Young’s modulus (same for same material), $F$ is same for all wires.

So:

$Δ L \propto \frac{L}{A} \propto \frac{L}{d^{2}}$

Step 2: Compute $L / d^{2}$ for each

(a) $L = 200 cm, d = 0.5 mm = 0.05 cm$

$\frac{L}{d^{2}} = \frac{200}{(0.05)^{2}} = \frac{200}{0.0025} = 80000$

(b) $L = 300 cm, d = 1.0 mm = 0.1 cm$

$\frac{L}{d^{2}} = \frac{300}{(0.1)^{2}} = \frac{300}{0.01} = 30000$

(c) $L = 50 cm, d = 0.05 mm = 0.005 cm$

$\frac{L}{d^{2}} = \frac{50}{(0.005)^{2}} = \frac{50}{0.000025} = 2,000,000$

(d) $L = 100 cm, d = 0.2 mm = 0.02 cm$

$\frac{L}{d^{2}} = \frac{100}{(0.02)^{2}} = \frac{100}{0.0004} = 250000$

The point at which an applied force A produces linear motion but no rotatory motion is _____?

Optical center
Mid-pint
Center of gravity
None of these

(0)

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Average distance of the earth from the sun A is L1. If one year of the earth =D days one year of another planet whose average distance from the sun is L2 will be ______?

D(L2/L1) days
D(L2/L1)1/2days
D(L2/L1)3/2 days
None of these

(0)

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Explanation

Step 1: Kepler’s third law
For planets around the Sun:

$T^{2} \propto a^{3}$

where $T$ is orbital period, $a$ is average distance from Sun.

Step 2: Apply to Earth and the other planet

$\frac{T_{2}}{T_{1}} = {(\frac{L_{2}}{L_{1}})}^{3 / 2}$

Given $T_{1} = D$ days (Earth’s year), $T_{2} = ?$

$T_{2} = D {(\frac{L_{2}}{L_{1}})}^{3 / 2}$

A particle is projected from the ground with C kinetic energy E at an angle of 60 degree with the horizontal. Its kinetic energy at the highest point of its motion will be ____?

E/2
E/4
E/8
None of these

(0)

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Explanation

Step 1: Initial velocity components
Let initial velocity = $u$ .
Initial kinetic energy $E = \frac{1}{2} m u^{2}$ .

At projection angle $60^{\circ}$ :
Horizontal component: $u_{x} = u \cos 60^{\circ} = \frac{u}{2}$
Vertical component: $u_{y} = u \sin 60^{\circ} = \frac{u \sqrt{3}}{2}$

Step 2: At highest point
Vertical velocity = 0, only horizontal velocity remains: $v = u_{x} = \frac{u}{2}$ .

Kinetic energy at highest point:

$E^{'} = \frac{1}{2} m {(\frac{u}{2})}^{2} = \frac{1}{2} m \frac{u^{2}}{4} = \frac{1}{4} (\frac{1}{2} m u^{2}) = \frac{1}{4} E$

PHYSICS

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

A progressive wave moving along X-axis is represented by y = Asin [2π/λ (vt- x)]. The wavelength (λ) at which the maximum particle velocity is 3 time the wave velocity is?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

22320 calheat is supplied to 100g of ice at zero degree centigrade. If the latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of water is 540 cal/g the final amount of water thus obtained and its temperature respectively are _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

An object placed in front of a concave mirror at a distance of X cm from the pole gives a 3 times magnified real image if it is moved to distance of X+5cm, the Magnification of the image becomes 2. The focal length of the mirror is _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

When a spring is stretched by 10 cm, the potential energy is stored is E. when the spring is stretched by 10cm more, the potential energy stored in the spring becomes?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A bullet on penetrating 30cm into its target A loses its velocity by 50%. What additional distance will its penetrate into the target before it comes to rest?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

Let L be the length and d be the diameter of B cross section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases the extension of wire will be the maximum?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

The point at which an applied force A produces linear motion but no rotatory motion is _____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

Average distance of the earth from the sun A is L1. If one year of the earth =D days one year of another planet whose average distance from the sun is L2 will be ______?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

A particle is projected from the ground with C kinetic energy E at an angle of 60 degree with the horizontal. Its kinetic energy at the highest point of its motion will be ____?

اس سوال کو وضاحت کے ساتھ پڑھیں

Explanation

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