Physics NCERT Exemplar Solutions Class 11th Chapter Seven

This is a multiple choice type question as classified in NCERT Exemplar

(b) The initial velocity is v_i = ve_y and after reflecting from the wall the final velocity is v_f= -ve_y. The trajectory is described as r= ye_y+ae_z.

hence the change in momentum is r $*m\left(v_f-v_i\right)=2mva{e}_x$  .

This is a multiple choice type question as classified in NCERT Exemplar

(c) Centre of mass lies towards the part of system of higher mass. In the above diagram the lower part is heavier bigger mass is downward . So COM lies below diameter.

This is a multiple choice type question as classified in NCERT Exemplar

(d) The bangle is in the form of ring. The centre of mass lies at the centre which is outside the body so C.O.M of bangle lies outside the circle.

This is a short answer type question as classified in NCERT Exemplar

Let b be the position vector of the centre of mass of a regular n-polygon.

(n-1) equal point masses are placed at (n-1) vertices of the regular n polygon, therefore for its centre of mass

r_CM= $\frac{\left(n-1\right)mb+ma}{\left(n-1\right)m+m}=0$

(n-1)mb+ma=0

B=- $\frac{a}{n-1}$

This is a short answer type question as classified in NCERT Exemplar

Where weight of the door acts along negative y axis

A force can produce torque only along a direction normal to itself as $\tau =r*F$ . So when the door is in the xy plane the torque produced by gravity can only along z direction. Never about an axis passing through y direction. Hence the weight will not produce any torque.

This is a short answer type question as classified in NCERT Exemplar

Wheel is a rigid body. The particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. This acceleration arises due to internal elastic forces which cancel out in pairs.

In a half wheel the distribution of mass about its centre of mass is not symmetrical, therefore the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence an external torque is required to maintain the motion of the wheel.

This is a short answer type question as classified in NCERT Exemplar

no $\sum _i{F}_i\ne 0$

The sum of torques about a certain point O $\sum _i{r}_i*F_i=0$

The sum of torques about any other O $\sum _i{r}_i{F}_i=0$

The sum of torques about any other point O'

$\sum _i(r_i-a)*F_i=\sum _i{r}_i*F_i-a*\sum _i{F}_i$

Here the second term need not vanish.

Sum of all torques about any point is zero.

This is a short answer type question as classified in NCERT Exemplar

Moment of force $\tau$ ₁= F (a)….anticlockwise

Moment of weight mg $\tau$ ₂=mg (a/2)……. (clockwise)

Cube will not exhibit motion then $\tau$ _{1 $=\tau$ 2}

$F*a=mg*\frac{a}{2}$

F=mg/2

Cube will rotate when $\tau$ _{1 $>\tau$ 2}

$F*a>mg*\frac{a}{2}$

F>mg/2

At a/3 from point A then

mg $*\frac{a}{3}=F*aorF=\frac{mg}{3}$

when F=mg/4 which is less then mg/3 there will be no motion.

This is a short answer type question as classified in NCERT Exemplar

As the slope of graph is positive and positive slope indicates anti clockwise rotation which is traditionally taken as positive.