Assertion (A): In projectile motion (from ground to ground projection), horizontal range is always same for angle of projection \(\theta\) and \(90^\circ – \theta\).
Reason (R): Horizontal range is independent of angle of projection.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): The range formula is \(R = \frac{u^2 sin(2\theta)}{g}\). For angle \((90^\circ - \theta)\), \(R' = \frac{u^2 sin(2(90^\circ - \theta))}{g} = \frac{u^2 sin(180^\circ - 2\theta)}{g} = \frac{u^2 sin(2\theta)}{g} = R\). So (A) is True.
Reason (R): The horizontal range clearly depends on the angle of projection (theta) via (sin(2theta)). So (R) is False.
Since (A) is true and (R) is false, option (3) is correct.
Assertion (A): In projectile motion, speed always decreases.
Reason (R): In presence of air drag, projectile motion is a uniformly accelerated motion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): In ideal projectile motion, speed decreases on the way up and increases on the way down, reaching a minimum at the peak. It does not always decrease. So (A) is False.
Reason (R): In the presence of air drag, the drag force depends on velocity, making the net acceleration non-constant. Thus, it is not uniformly accelerated motion. So (R) is False.
Since both (A) and (R) are false, option (4) is correct.
Assertion (A): When speed of projection of a body is made (n) times, its time of flight becomes (n) times.
Reason (R): At this speed, the range of projectile becomes (n^2) times.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Time of flight \(T = \frac{2u sin\theta}{g}\). If (u) is replaced by (nu), (T' = nT). So (A) is True.
Reason (R): Horizontal range \(R = \frac{u^2 sin(2\theta)}{g}\). If (u) is replaced by (nu), (R' = n^2 R). So (R) is True.
Both statements are true. However, the scaling of range (R) does not explain the scaling of time of flight (A). They are independent consequences of initial speed scaling. So (R) is not the correct explanation for (A). Option (2) is correct.
Assertion (A): When the range of a projectile is maximum, the time of flight is the largest.
Reason (R): Horizontal range is maximum when angle of projection is (90^circ).
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Maximum range occurs at (theta = 45^circ). The largest time of flight occurs at (theta = 90^circ). Since these angles are different, the assertion is false. So (A) is False.
Reason (R): Horizontal range is maximum at (theta = 45^circ) (when (sin(2theta)=1)). At (theta = 90^circ), the range is zero. So (R) is False.
Since both (A) and (R) are false, option (4) is correct.
Assertion (A): A particle has positive acceleration it means that its speed always increases.
Reason (R): Acceleration is the rate of change of speed with respect to time.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is false because positive acceleration doesn't always mean increasing speed; it depends on the direction of velocity. Speed increases only when \(vec{a}\) and \(vec{v}\) are in the same direction. Reason (R) is false because acceleration is the rate of change of velocity, not speed.
Assertion (A): Trajectory of an object moving under a constant acceleration must be a straight line.
Reason (R): The shape of trajectory depends only on the acceleration.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is false; under constant acceleration, the trajectory can be a parabola (projectile motion) or a straight line depending on initial velocity. Reason (R) is false; the trajectory's shape depends on both initial velocity and acceleration.
A particle begins to move along straight line where the acceleration \( (a) \) of the particle varies with displacement \( (x) \) according to relation, \( a = 5x \), then velocity of the particle varies with displacement as
1. \( x^{1/2} \)
2. \( x^1 \)
3. \( x^{1/3} \)
4. \( x^{3/4} \)
View Answer
Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).
A car starts from rest, accelerates uniformly at \( 2 \, \text{m/s}^2 \). The distance travelled by the car in fourth second is
1. 2.25 m
2. 14 m
3. 7 m
4. Zero
View Answer
The distance travelled in the \( n^{\text{th}} \) second is \( s_n = u + \frac{a}{2}(2n - 1) \). Substituting \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( n = 4 \) yields \( s_4 = 0 + \frac{2}{2}(2(4) - 1) = 7 \, \text{m} \).
A particle is projected with a speed of \( 20 \, \text{m s}^{-1} \) from level ground at an angle \( \theta \) equal to \( 45^\circ \) from horizontal. The ratio of maximum height attained by the body to horizontal range acquired by the body will be
1. \( \frac{2}{1} \)
2. \( \frac{1}{2} \)
3. \( \frac{1}{4} \)
4. \( \frac{3}{2} \)
View Answer
The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).