Question 61:
A particle moves in x-y plane according to rule \(x = a sin omega ttext{ and }y = a cos omega ttext{. The particle follows: (2010 Mains)}\
Concept: Parametric equations of motion define the trajectory.
Formula: \(sin^2 theta + cos^2 theta = 1\).
Given \(x = a sin omega ttext{ and }y = a cos omega t\).
Square and add: \(x^2 + y^2 = (a sin omega t)^2 + (a cos omega t)^2 = a^2 (sin^2 omega t + cos^2 omega t) = a^2\).
This is the equation of a circle \(x^2 + y^2 = a^2\).
Question 62:
A body is whirled in a horizontal circle of radius \( 20 text{ cm} \) with an angular velocity of \( 10 text{ rad/s} \). What is its linear velocity at any point on its circular path? (1996)
Given: Radius \( r = 20 text{ cm} = 0.20 text{ m} \). Angular velocity \( omega = 10 text{ rad/s} \). The relationship between linear velocity \( v \) and angular velocity \( omega \) is \( v = romega \). Substituting the values, \( v = (0.20 text{ m}) times (10 text{ rad/s}) = 2 text{ m/s} \).
Question 63:
The angular speed of a flywheel making \( 120 text{ revolution/minute} \) is: (1995)
Given: Frequency \( f = 120 text{ revolution/minute} \). Convert to revolutions per second: \( f = frac{120}{60} = 2 text{ rev/s} \). The angular speed is given by \( omega = 2pi f \). Substituting the frequency, \( omega = 2pi (2) = 4pi text{ rad/s} \).
Question 64:
An electric fan has blades of length \( 30 text{ cm} \) measured from the axis of rotation. If the fan is rotating at \( 120 text{ rpm} \), the acceleration of a point on the tip of the blade is: (1990)
Given: Radius \( r = 30 text{ cm} = 0.30 text{ m} \). Frequency \( f = 120 text{ rpm} = frac{120}{60} = 2 text{ rps} \). Angular velocity \( omega = 2pi f = 2pi (2) = 4pi text{ rad/s} \). The centripetal acceleration is \( a_c = romega^2 \). Thus, \( a_c = 0.30 times (4pi)^2 = 0.30 times 16pi^2 approx 0.30 times 16 times (3.14159)^2 approx 47.37 text{ m s}^{-2} \). Approximately \( 47.4 text{ m s}^{-2} \).
Question 65:
A particle moves along a circle of radius \( left( frac{20}{pi} right) text{ m} \) with constant tangential acceleration. If the velocity of the particle is \( 80 text{ m/s} \) at the end of the second revolution after motion has begun, the tangential acceleration is: (2003)
Given: \( r = frac{20}{pi} text{ m} \), \( v = 80 text{ m/s} \). The angular displacement is \( theta = 2 times 2pi = 4pi text{ rad} \). The distance covered is \( s = rtheta = frac{20}{pi} times 4pi = 80 text{ m} \). Using the kinematic equation \( v^2 = u^2 + 2as \), we get \( (80)^2 = 0 + 2 a_t (80) \), which yields \( a_t = 40 text{ m/s}^2 \).
Question 66:
Two particles having mass \( ‘M’ \) and \( ‘m’ \) are moving in a circular path having radius \( R \) and \( r \) respectively. If their time period are same then the ratio of angular velocity will be: (2001)
Angular velocity is defined as \( omega = frac{2pi}{T} \). Since the time period \( T \) is the same for both particles, their angular velocities will be equal. Therefore, the ratio of their angular velocities \( frac{omega_1}{omega_2} = frac{2pi/T}{2pi/T} = 1 \).
Question 67:
A small block slides down on a smooth inclined plane, starting from rest at time \(t=0\). Let \(S_n\) be the distance travelled by the block in the interval \(t=n-1\) to \(t=n\). The, the ratio \(frac{S_n}{S_{n+1}}\) is: (2021)
For a body starting from rest with constant acceleration \(a\), the distance traveled in the \(n^{text{th}}\) second is \(S_n = frac{1}{2} a (2n - 1)\). Similarly, \(S_{n+1} = frac{1}{2} a (2(n+1) - 1) = frac{1}{2} a (2n + 1)\). The ratio is \(frac{S_n}{S_{n+1}} = frac{frac{1}{2} a (2n - 1)}{frac{1}{2} a (2n + 1)} = frac{2n - 1}{2n + 1}\).
Question 68:
A balloon with mass \(m\) is descending down with an acceleration \(a\) (where \(a < g\)). How much mass should be removed from it so that it starts moving up with an acceleration \(a\)? (2014)
Let buoyant force be \(F_B\). Descending: \(mg - F_B = ma Rightarrow F_B = m(g-a)\). To ascend with acceleration \(a\), let mass removed be \(m'\). Then \(F_B - (m-m')g = (m-m')a\). Substitute \(F_B\): \(m(g-a) - (m-m')g = (m-m')a\). Solving for \(m'\), we get \(m' = frac{2ma}{g+a}\).
Question 69:
A body, under the action of a force \(vec{F} = 6hat{i} – 8hat{j} + 10hat{k}\) acquires an acceleration of \(1text{ m/s}^2\). The mass of this body must be: (2009, 1996)
The magnitude of the force is \(|vec{F}| = sqrt{6^2 + (-8)^2 + 10^2} = sqrt{36+64+100} = sqrt{200} = 10sqrt{2}text{ N}\). According to Newton's second law, \(F = ma\). Given \(a = 1text{ m/s}^2\), so \(m = F/a = frac{10sqrt{2}text{ N}}{1text{ m/s}^2} = 10sqrt{2}text{ kg}\).
Question 70:
A man weighs \(80text{ kg}\) . He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of \(5text{ m/s}^2\). What would be the reading on the scale? \((g = 10text{ m/s}^2)\): (2003)
When a lift accelerates upwards, the apparent weight \(R\) is given by \(R = m(g+a)\). Given \(m = 80text{ kg}\), \(g = 10text{ m/s}^2\), and \(a = 5text{ m/s}^2\). Therefore, \(R = 80text{ kg} times (10text{ m/s}^2 + 5text{ m/s}^2) = 80 times 15 = 1200text{ N}\).