A car of mass \(1000\) kg negotiates a banked curve of radius \(90\) m on a frictionless road. If the banking angle is of \(45^\circ,\) the speed of the car is:
1. | \(20\) ms–1 | 2. | \(30\) ms–1 |
3. | \(5\) ms–1 | 4. | \(10\) ms–1 |
A particle moving with velocity \(\vec{v}\) is acted by three forces shown by the vector triangle \(\mathrm{PQR}.\) The velocity of the particle will:
1. | change according to the smallest force \(\mathrm{\overrightarrow{Q R}}\) |
2. | increase |
3. | decrease |
4. | remain constant |
Two stones of masses \(m\) and \(2m\) are whirled in horizontal circles, the heavier one in a radius \(\frac{r}{2}\) and the lighter one in a radius \(r\). The tangential speed of lighter stone is \(n\) times that of the value of heavier stone when they experience the same centripetal forces. The value of \(n\) is:
1. | \(3\) | 2. | \(4\) |
3. | \(1\) | 4. | \(2\) |
A car is negotiating a curved road of radius \(R\). The road is banked at an angle \(\theta\). The coefficient of friction between the tyre of the car and the road is \(\mu_s\). The maximum safe velocity on this road is:
1. \(\sqrt{\operatorname{gR}\left(\frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}\right)}\)
2. \(\sqrt{\frac{\mathrm{g}}{\mathrm{R}}\left(\frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}\right)}\)
3. \(\sqrt{\frac{\mathrm{g}}{\mathrm{R}^2}\left(\frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\operatorname{s}} \tan \theta}\right)}\)
4. \(\sqrt{\mathrm{gR}^2\left(\frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}\right)}\)
One end of the string of length \(l\) is connected to a particle of mass \(m\) and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed \(v\), the net force on the particle (directed towards the center) will be: (\(T\) represents the tension in the string)
1. \(T+\frac{m v^2}{l}\)
2. \(T-\frac{m v^2}{l}\)
3. zero
4. \(T\)
A massless and inextensible string connects two blocks \(\mathrm{A}\) and \(\mathrm{B}\) of masses \(3m\) and \(m,\) respectively. The whole system is suspended by a massless spring, as shown in the figure. The magnitudes of acceleration of \(\mathrm{A}\) and \(\mathrm{B}\) immediately after the string is cut, are respectively:
1. | \(\frac{g}{3},g\) | 2. | \(g,g\) |
3. | \(\frac{g}{3},\frac{g}{3}\) | 4. | \(g,\frac{g}{3}\) |
Conservation of momentum in a collision between particles can be understood from:
1. | conservation of energy |
2. | newton's first law only |
3. | newton's second law only |
4. | both Newton's second and third law |
Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly and the surface is \(0.05\). ( \(g=10 \mathrm{~m} / \mathrm{s}^2\), mass of the string is negligible and no other friction exists ).
1. | \( 1.25 \mathrm{~m} / \mathrm{s}^2 \) | 2. | \( 1.50 \mathrm{~m} / \mathrm{s}^2 \) |
3. | \( 1.66 \mathrm{~m} / \mathrm{s}^2 \) | 4. | \( 1.00 \mathrm{~m} / \mathrm{s}^2\) |
A cricketer catches a ball of mass \(150~\mathrm{gm}\) in \(0.1\) \(\mathrm{s}\) moving with a speed of \(20~\mathrm{ms^{-1}}\). Then he experiences a force of:
1. \(300~\mathrm{N}\)
2. \(30~\mathrm{N}\)
3. \(3~\mathrm{N}\)
4. \(0.3~\mathrm{N}\)
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12\) m/s. If the mass of the ball is \(0.15\) kg, then the impulse imparted to the ball is:
(Assume linear motion of the ball.)
1. \(0.15\) N-s
2. \(3.6\) N-s
3. \(36\) N-s
4. \(0.36\) N-s