HSSC-I (Class-11) Solved Physics MCQs; Unit-3: Force and Motion (Baluchistan Textbook Board, Quetta) With Verified Answers:
The shortest distance is called displacement. 4m + 3m = 7m
2. Acceleration due to the uniform velocity of a body is
If a body is moving with uniform velocity, the acceleration will be zero in the direction of motion.
3. Third equation of motion is independent of:
2aS= vf² - vi²
S = 80-20 = 60 m and t = 4 hrs - 2 hrs = 2, so, v = `\frac {s}{t}` = v = `\frac {60 m}{2 h}` = 30 km h⁻¹
5. What will be the velocity of a body when it starts its motion from rest and after 5s its acceleration becomes 2 m s⁻¹
vi = 0 m s⁻¹, a = 2 m s⁻¹, t = 5s, Now using 1st equation of motion vf = vi + at = 0 + 2 x 5 = 10 m s⁻¹
Velocity = displacement / time `\vec v` = `\frac {vec v}{t}` Here displacemennt is two radius ie 2 x 2m = 4m and t= 2s so, `\vec v` = `\frac {4m}{2s}`= 2 m s⁻¹
vi = 0 m s⁻¹, a = 2 m s⁻¹, t = 5s, Now using 1st equation of motion vf = vi + at = 0 + 2 x 5 = 10 m s⁻¹
6. What is the velocity of an object when it reaches point P to point Q in 2s along a semicircle radius of 2m?
Velocity = displacement / time `\vec v` = `\frac {vec v}{t}` Here displacemennt is two radius ie 2 x 2m = 4m and t= 2s so, `\vec v` = `\frac {4m}{2s}`= 2 m s⁻¹
7. The Sl unit of weight is :
Weight is force
8. What is the acceleration produced by a force of 0.5 N applied on a body of mass 0.1 kg?
F= ma or a = `\frac {F}{m}` = `\frac {0.5 N}{0.1 kg}` = 5 m s⁻²
9. Three masses that are connected by a massless spring are shown in the figure. What will be the value of F₂?
⇒ F₁ = 5a kg -----------(1) where a is the acceleration ⇒ F₂ - F₁ = 3a kg --------(2) ⇒ 30 N - F₂ = 2a kg --------(3) ⇒ by adding (1) and (2) we get F₂ = 8a kg ------------(4) ⇒ by adding (3) and (4) we get 30 N = 10 a kg or a = 3 m s⁻² ⇒ putting the value of an equation (4) F₂ = 8 (3 m s⁻²) kg = 24 N
10. Rate of change of momentum is
F = ma = m `\frac{v}{t}` =`\frac{mv}{t}` = `\frac{ΔP}{Δt}`
11. A body moves from point P to Q with a speed of 6 m s⁻¹ along a straight line then from Q to P with a speed of 4 m s⁻¹. What is the average speed over the entire trip?
Average Speed = `\frac {v_1 + v_2}{2}`
Impulse = ΔF x Δt
13. Impulse always changed
Impulse-Momentum Theorem
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As the horizontal component of the velocity remain constant in projectile motion, so ax = 0
Formula for Height of Projectile H = `\frac {(v_0 sinፀ)^2 }{2g}` so, if ፀ = 90° and Sine 90° =1 so maximum height
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Compare the equation of Range and height of the projectile and we will get 0 = tan⁻¹ (4) = 76°
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Impulse-Momentum Theorem
14. In a one-dimensional elastic collision of two bodies of the same masses, what will happen if the moving body collides with the mass which is initially at rest?
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15. In projectile motion, the horizontal component of acceleration of a body is:
As the horizontal component of the velocity remain constant in projectile motion, so ax = 0
16.. At what angle does a projectile gain its maximum height:
Formula for Height of Projectile H = `\frac {(v_0 sinፀ)^2 }{2g}` so, if ፀ = 90° and Sine 90° =1 so maximum height
17. A rocket works on basis of :
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18. The vertical and horizontal distances of the projectile will be equal if the angle of projection is:
Compare the equation of Range and height of the projectile and we will get 0 = tan⁻¹ (4) = 76°
19. The quantitative measure of the inertia of a body is:
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