Question 51:
The maximum range of a gun of horizontal terrain is \(16 text{ km}\). If \(g = 10 text{ m s}textsuperscript{-2}\), then muzzle velocity of a shell must be: (1990)
Concept: Maximum range of a projectile.
Formula: Maximum range \(R_{max} = u^2 / g\) (at \(45°\)).
Given \(R_{max} = 16 times 10^3text{ m}\), \(g = 10text{ m/s}textsuperscript{2}\).
\(u = sqrt{R_{max} cdot g} = sqrt{16 times 10^3 times 10} = sqrt{16 times 10^4} = 400text{ m/s}\).
Question 52:
The speed of a swimmer in still water is \(20 text{ m/s}\). The speed of river water is \(10 text{ m/s}text{ and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by: (2019)}\
Concept: Relative velocity for shortest path across a river.
Formula: \(sin theta = v_r / v_s\), where \(theta\) is the angle upstream.
Given \(v_s = 20text{ m/s}\), \(v_r = 10text{ m/s}\).
For the shortest path, \(sin theta = 10/20 = 1/2\). Thus, \(theta = 30°\).
The direction is \(30°text{ west of north}\).
Question 53:
A ship A is moving Westwards with a speed of \(10 text{ km/h}text{ and a ship B 100 km South of A, is moving Northwards with a speed of 10 km/h. The time after which the distance between them becomes shortest, is: (2015)}\
Concept: Relative velocity for minimum distance between moving objects.
Formula: \(t = -(vec{r_{rel}} cdot vec{v_{rel}}) / v_{rel}^2\).
\(vec{v_A} = -10 hat{i}\), \(vec{v_B} = 10 hat{j}\). \(vec{v_{BA}} = 10 hat{i} + 10 hat{j}\).
Initial relative position \(vec{r_{BA}}(0) = -100 hat{j}\).
\(t = -((-100 hat{j}) cdot (10 hat{i} + 10 hat{j})) / (10^2 + 10^2) = -(-1000) / 200 = 5 text{ h}\).
Question 54:
The width of river is \(1 text{ km}\). The velocity of boat is \(5 text{ km/hr}\). The boat covered the width of river with shortest possible path in \(15 text{ min}\). Then the velocity of river stream is: (2000)
Concept: Relative velocity for crossing a river along the shortest path.
Formula: \(v_g = d/t\), \(v_b^2 = v_g^2 + v_r^2\).
Resultant velocity \(v_g = (1text{ km}) / (1/4text{ h}) = 4text{ km/h}\).
Using \(v_b^2 = v_g^2 + v_r^2\), we get \(5^2 = 4^2 + v_r^2\).
So, \(v_r = sqrt{25 - 16} = sqrt{9} = 3text{ km/h}\).
Question 55:
The speed of a boat is \(5 text{ km/hr}text{ in still water. It crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of river water is: (1998)}\
Concept: Relative velocity for crossing a river along the shortest path.
Formula: \(v_g = d/t\), \(v_b^2 = v_g^2 + v_r^2\).
Resultant velocity \(v_g = (1text{ km}) / (1/4text{ h}) = 4text{ km/h}\).
Using \(v_b^2 = v_g^2 + v_r^2\), we get \(5^2 = 4^2 + v_r^2\).
So, \(v_r = sqrt{25 - 16} = sqrt{9} = 3text{ km/h}\).
Question 56:
A boat is sent across a river with a velocity of \(8 text{ km h}textsuperscript{-1}\). If the resultant velocity of boat is \(10 text{ km h}textsuperscript{-1}\), then velocity of river is: (1994, 93)
Concept: Resultant velocity in river-boat problems.
Formula: \(vec{v}_g = vec{v}_b + vec{v}_r\). If \(vec{v}_b\) is perpendicular to \(vec{v}_r\), then \(v_g^2 = v_b^2 + v_r^2\).
Given \(v_b = 8text{ km/h}\), \(v_g = 10text{ km/h}\).
\(v_r = sqrt{v_g^2 - v_b^2} = sqrt{10^2 - 8^2} = sqrt{100 - 64} = sqrt{36} = 6text{ km/h}\).
Question 57:
A bus is moving on a straight road towards north with a uniform speed of \(50 text{ km/hour}text{ then it turns left through 90°. If the speed remains unchanged after turning, the increase in the velocity of bus in the turning process is: (1989)}\
Concept: Change in velocity is a vector difference.
Formula: \(Delta vec{v} = vec{v}_f - vec{v}_i\).
Initial velocity \(vec{v}_i = 50hat{j}\). Final velocity \(vec{v}_f = -50hat{i}\).
\(Delta vec{v} = -50hat{i} - 50hat{j}\). Magnitude \(|Delta vec{v}| = sqrt{(-50)^2 + (-50)^2} = 50sqrt{2} approx 70.7text{ km/hr}\).
Direction: South-West.
Question 58:
Two particles A and B are moving in uniform circular motion in concentric circles of radii \(r_A\) and \(r_B\) with speed \(v_A\) and \(v_B\) respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be: (2019)
Concept: Relationship between angular speed and time period in UCM.
Formula: Angular speed \(omega = 2pi / T\).
Given that time period \(Ttext{ is same for both A and B}\).
Therefore, \(omega_A = 2pi / T\) and \(omega_B = 2pi / T\).
Thus, \(omega_A = omega_B\), and the ratio \(omega_A : omega_B = 1:1\).
Question 59:
When an object is shot from the bottom of a long smooth inclined plane kept at an angle \(60°text{ with horizontal, it can travel a distance }x_1text{ along the plane. But when the inclination is decreased to }30°text{ and the same object is shot with the same velocity, it can travel }x_2text{ distance. Then }x_1text{ : }x_2text{ will be: (2019)}\
Concept: Motion on an inclined plane with gravity.
Formula: Distance traveled up the incline \(x = u^2 / (2g sin alpha)\).
For \(alpha_1 = 60°\), \(x_1 = u^2 / (2g sin 60°) = u^2 / (gsqrt{3})\).
For \(alpha_2 = 30°\), \(x_2 = u^2 / (2g sin 30°) = u^2 / g\).
Ratio \(x_1 : x_2 = (u^2 / (gsqrt{3})) : (u^2 / g) = 1 : sqrt{3}\).
Question 60:
A particle moves in a circle of radius \(5 text{ cm}text{ with constant speed and time period }0.2pitext{ s}text{. The acceleration of the particle is (2011 Pre)}\
Concept: Centripetal acceleration in uniform circular motion.
Formula: \(a_c = omega^2 r = (2pi / T)^2 r\).
Given \(r = 0.05text{ m}\), \(T = 0.2pi text{ s}\).
\(a_c = (2pi / (0.2pi))^2 times 0.05 = (10)^2 times 0.05 = 100 times 0.05 = 5text{ m/s}textsuperscript{2}\)