A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be 

1. $mg \cos \theta$

2. $mg \sin \theta$

3. mg

4. $mg/\cos \theta$

A particle moves in the \(XY\text-\)plane under the action of a force \(F\) such that the components of its linear momentum \(p\) at any time \(t\) are \(p_x = 2 \cos t\),  \(p_y = 2 \sin t\). The angle between \(F\) and \(p\) at time \(t\) will be:

On a stationary sailboat, the air is blown at the sails from a fan attached to the boat. The boat will:

1. remain stationary

2. spin around

3. move in a direction opposite to that in which air is blown

4. move in the direction in which the air is blown

The tension in the spring is 

1. Zero

2. 2.5 N

3. 5 N

4. 10 N

Rocket engines lift a rocket from the earth surface because hot gases with high velocity:

1. push against the earth.

2. push against the air.

3. react against the rocket and push it up.

4. heat up the air which lifts the rocket.

The mass of a body measured by a physical balance in a lift at rest is found to be m. If the lift is going up with an acceleration a, its mass will be measured as:

1. $m\left(1-\frac{{\displaystyle a}}{{\displaystyle g}}\right)$

2. $m\left(1+\frac{{\displaystyle a}}{{\displaystyle g}}\right)$

3. m

4. Zero

Three weights W, 2W and 3W are connected to identical springs suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weights from the rod are such that 

1. 3W will be farthest

2. W will be farthest

3. All will be at the same distance

4. 2W will be farthest

Consider the following statement: When jumping from some height, you should bend your knees as you come to rest, instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?

1. $\Delta \overrightarrow{p_1}=-\Delta \overrightarrow{p_2}$

2. $\Delta E=-\Delta (PE+KE)=0$

3. $\overrightarrow{F}\Delta t=m\Delta \overrightarrow{v}$

4. $\Delta \overrightarrow{x}\propto \Delta \overrightarrow{F}$

In the arrangement shown in the figure, the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed 

1. $2U \cos \theta$

2. $U \cos \theta$

3. $\frac{2U}{\cos \theta }$

4. $\frac{U}{\cos \theta }$

A string of negligible mass going over a clamped pulley of mass \(m\) supports a block of mass \(M\) as shown in the figure. The force on the pulley by the clamp is given by:
 
1. \(\sqrt{2} M g\)
2. \(\sqrt{2} m g\)
3. \(g\sqrt{\left( M + m \right)^{2} + m^{2}}\)
4. \(g\sqrt{\left(M + m \right)^{2} + M^{2}}\)