Q. No.A particle is projected with a speed of \(2~\text{m/s}\) from the base of a plane inclined at \(30^\circ\) to the horizontal. The direction of projection makes an angle of \(15^\circ\) above the inclined plane as shown in the figure. If \(g=10~\text{m/s}^2,\) what is the distance along the plane from the point of projection to the point where the particle strikes the plane?
1.
\(14~\text{cm}\)
2.
\(28~\text{cm}\)
3.
\(20~\text{cm}\)
4.
\(36~\text{cm}\)
Two particles are projected from the same point with the same speed \(u,\) but at different angles. They both cover the same horizontal range \(R,\) but reach different maximum heights, \(h_1\) and \(h_2.\) Which of the following relations is correct?
| 1. | \(R^2=h_1h_2\) | 2. | \(R^2=16h_1h_2\) |
| 3. | \(R^2=4h_1h_2\) | 4. | \(R^2=2h_1h_2\) |
A shell is fired from a fixed artillery gun with an initial speed \(u\) such that it hits the target on the ground at a distance \(R\) from it. If \(t_1\) and \(t_2\) are the values of the time taken by it to hit the target in two possible ways, the product \(t_1t_2\) is:
1. \(\frac{2R}{g}\)
2. \(\frac{R}{2g}\)
3. \(\frac{R}{g}\)
4. \(\frac{R}{4g}\)
The trajectory of a projectile near the surface of the earth is given as \(y=2x-9x^2\). If it were launched at an angle \(\theta_0\) with speed \(v_0\) then:\(\left(g= 10~\text{ms}^{-2}\right)\)
| 1. | \(\theta_0=\cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \text{ and }v_0=\frac{5}{3}~\text{ms}^{-1}\) |
| 2. | \(\theta_0=\cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \text{ and }v_0=\frac{3}{5}~\text{ms}^{-1}\) |
| 3. | \(\theta_0=\sin^{-1}\left(\frac{2}{\sqrt{5}}\right) \text{ and }v_0=\frac{3}{5}~\text{ms}^{-1}\) |
| 4. | \(\theta_0=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \text{ and }v_0=\frac{5}{3}~\text{ms}^{-1}\) |
\(\vec{r}(t)=(15 t^2) \hat{i}+\left(4-20 t^2\right) \hat{j},\) where \(\vec{r}(t)\) is in metres and \(t\) is in seconds.
What is the magnitude of the acceleration at \(t=1\) second?
| 1. | \(100\) m/s2 | 2. | \(40\) m/s2 |
| 3. | \(50\) m/s2 | 4. | \(25\) m/s2 |
A helicopter rises from rest on the ground vertically upwards with a constant acceleration \(g\). A food packet is dropped from the helicopter when it is a height \(h\). The time taken by the packet to reach the ground is close to [\(g\) is the acceleration due to gravity]:
1. \( t=\sqrt{\frac{2 h}{3 g}} \)
2. \( t=1.8 \sqrt{\frac{h}{g}} \)
3. \( t=3.4 \sqrt{\left(\frac{h}{g}\right)} \)
4. \( t=\frac{2}{3} \sqrt{\left(\frac{h}{g}\right)}\)
A particle starts from the origin at time \(t=0 \) with an initial velocity of \(5\hat{j}~\text{ms}^{-1}. \) It moves in the \(XY \text-\)plane under a constant acceleration of \(\left(10\hat{i}+4\hat{j}\right)~\text{ms}^{-2} .\) At some later time \(t,\) the coordinates of the particle are \((20~\text{m}, y_0~\text{m}). \) The values of \(t \) and \(y_0 \) are, respectively:
1. \(4~\text{s}\) and \(52~\text{m}\)
2. \(5~\text{s}\) and \(25~\text{m}\)
3. \(2~\text{s}\) and \(18~\text{m}\)
4. \(2~\text{s}\) and \(24~\text{m}\)
A particle moves at a constant speed along the circumference of a circle with a radius \(R,\) subject to a central fictitious force \(F\) that is inversely proportional to \(R^3.\) Its time period of revolution will be given by:
| 1. | \( T \propto R^2 \) | 2. | \( T \propto R^{\frac{3}{2}} \) |
| 3. | \( T \propto R^{\frac{5}{2}} \) | 4. | \(T \propto R^{\frac{4}{3}} \) |
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed \(v\), he sees that rain drops are coming at an angle \(60^\circ\) from the horizontal. On further increasing the speed of the car to \((1+\beta)v\), this angle changes to \(45^\circ\). The value of \(\beta\) is close to:
1. \(0.41\)
2. \(0.50\)
3. \(0.37\)
4. \(0.73\)
A clock has a continuously moving second's hand of \(0.1~\text{m}\) length. The average acceleration of the tip of the second hand (in units of ms-2) is of the order of:
1. \(10^{-3}\)
2. \(10^{-4}\)
3. \(10^{-1}\)
4. \(10^{-2}\)
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