PHYSICS LECTURER PAST PAPER 20-01-2025
Lecturer Physics 5 years Past Papers Click Here
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چار آپشن میں سے کسی ایک پر کلک کرنے سے جواب سرخ ہو جائے گا۔
جب جسم پر لگائے جانے والے تناؤ کو زیادہ سے زیادہ قیمت سے زیادہ بڑھایا جاتا ہے اور کچھ وقت کے بعد اسے ہٹا دیا جاتا ہے تو کیا ہوتا ہے؟ ینگ کے ماڈیولس کی تعریف ______ کے طور پر کی گئی ہے؟
Tensile stress/tensile strain
Tensile strain/tensile stress
Tensile strain × tensile stress
None of these
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Explanation
Young's modulus is a measure of the stiffness of a material, defined as the ratio of tensile stress to tensile strain within the elastic limit. It quantifies how much a material will stretch or compress under a given force.
65
80
89
None of these
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Explanation
To find the divergence, calculate:
div(f) = ∂/∂x (3x^2) + ∂/∂y (5xy) + ∂/∂z (xyz^3)
= 6x + 5x + 3xyz^2
At (1, 2, 3):
div(f) = 6(1) + 5(1) + 3(1)(2)(3)^2
= 6 + 5 + 54
= 65
بوہر کی مقدار کی حالت _____ ہے؟
mvr = n^2h/π
mvr = h/2π
mvr = nh/2π
None of these
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Explanation
Bohr's quantum condition states that the angular momentum of an electron in a circular orbit is quantized:
mvr = nh / 2π
where m is the mass of the electron,
v is the velocity ,
r is the radius of the orbit,
n is the principal quantum number,
and h is Planck's constant .
V/R^2
V^2/R
Zero
None of these
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Explanation
The acceleration of a particle moving in a uniform circular motion is directed towards the center of the circle and is given by:
a = v^2 / r
This is known as centripetal acceleration.
2/7
3/7
1/7
None of these
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Explanation
When a spherical ball rolls without slipping:
Total energy (E) = Kinetic energy (K) + Rotational energy (U)
K = (1/2)mv^2
U = (1/2)Iω^2 = (1/5)mv^2 (for a sphere)
E = K + U = (1/2)mv^2 + (1/5)mv^2 = (7/10)mv^2
Fraction of energy associated with rotation:
U/E = ((1/5)mv^2) / ((7/10)mv^2) = 2/7
4 F0
16 F0
2 F0
None of these
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Explanation
According to the law of universal gravitation:
F ∝ (m1 × m2) / r^2
Initially, F0 ∝ (m × m) / r^2
After doubling the mass and halving the distance:
F_new ∝ (2m × 2m) / (r/2)^2
= 16 × F0
2.45 X 10^5 Nm^2/C
1.12 X 10^5 Nm^2/C
3.12 X 10^5 Nm^2/C
None of these
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Explanation
We know that for a Gaussian surface, flux = q/ε
Here, q = 10^-6 C and ε = 8.85 X 10^-12 C/Nm^2
Therefore, flux = 10^-6/8.85 X 10^-12
= 1.12 X 10^5 Nm^2/C.
2.4961/m^3
4524.2×10^18/m^3
2.4961×10^18/m^3
None of these
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Explanation
Energy of one photon (E) = hc / λ
h = 6.626 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 4961 Å = 4961 × 10^-10 m
E = (6.626 × 10^-34 Js × 3 × 10^8 m/s) / (4961 × 10^-10 m)
= 4.007 × 10^-19 J
Number of photons required to do 1 J of work:
n = 1 J / (4.007 × 10^-19 J)
= 2.4961 × 10^18
8.4 × 10^-3 J
8.0 × 10^-3 J
8.2 × 10^-3 J
None of these
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Explanation
Total mechanical energy (E) = (1/2) × m × ω^2 × A^2
ω = 2π / T = 2π / 0.20 s = 10π rad/s
m = 40 g = 0.04 kg
A = 2.0 cm = 0.02 m
E = (1/2) × 0.04 kg × (10π rad/s)^2 × (0.02 m)^2
= 8.0 × 10^-3 J
T=2π√m/k1+k2
T=2π√m/k1
T=2π√k1+k2/m
None of these
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Explanation
When two springs are connected in parallel: Effective spring constant (k) = k1 + k2
Time period (T) = 2π√(m / k)
= 2π√(m / (k1 + k2))