Learning with Gyanendra Singh: Tutoring Sessions

Answer

Let AB be the rod of mass M and length L at an angle θ with the axis of rotation z.

Assume an infinitesimally small piece of length dx at a distance x from A as shown in the figure.

To find the moment of inertia I of the rod about the axis of rotation, we need to find the moment of inertia dI (about the axis of rotation) of this small piece of length in terms of variable x & other given constants and integrate the result over the length AB of the rod.

Linear density of the rod ρ = M/L

Mass of the small piece dm = ρ dx = (M/L)dx

dI = dm (r2) = (M/L)dx (x cos θ)2

dI = [(M cos2 θ)/L] x2 dx

 Integrating with respect to x within limits 0 to L

I = (1/3) ML2 cos2 θ

Answer

Correcct answer is A

Suppose a person with mass m lands on the net with velocity v and bounces back to height h after landing on the net. Assuming no loss of energy during the landing and bouncing back

(½)mv2 = mgh

h = v2/2g     i.e. independent of mass of the person

Also, landing velocities of Sara and Josh will be the same (i.e. 20 m/s, independent of mass)

So, h is independent of m and will be the same for Sara and Josh as they landed on the net with same velocity. Hence statement A is Not true.

B is true. (u = 0, v = 20 m/s)

C is true (Same acceleration, force is dependent on mass. Josh has higher mass hence will experience higher force)

D is true (Potential energy of the body converted to Kintic energy converted to elastic energy of the net converted to Kinetic energy of the body)

Answer

The actual value of coefficient of linear expansion of copper is 1.7 x 10^(-5) ̊C^(-1)             [Not 10^(-50)]

So coefficient of areal expansion = 3.4 x 10^(-5)

New area = 250 * [1 + 3.4 x 10^(-5) x (60-0)]

New area = 250 * 1.00204 = 250.51 cm^2

Answer

Note: Although it is written that Gravitational force is proportional to R^(5/2), we must assume it is inversely proportional to R^(5/2) otherwise it would mean that the Gravitational force is increasing with distance between the stars which is not possible.

Let the mass of the massive star be M and that of the small star is m.

The centrifugal force required to keep the star in orbit is provided by the gravitational force between the star and the massive star

i.e. (G*M*m)/R^(5/2) = (m*V^2)/R

So, V^2 = (G*M)/R^(3/2)

Now T = 2πR/V

Or T^2 = (4π^2)*(R^2)/(V^2) = [(4π^2)/GM]*R^(7/2)

i.e. T^2 is directly proportional to R^(7/2)

Hence B is the correct answer

Answer

Magnitude of change in velocity is given by 

V^2 + V^2 - 2*V*V*cos θ

= 2*V^2 * (1 - cos θ)

= 4*V^2 * sin ^2 (θ/2)

Answer

Correct answer is C. Please see the attached image for explanation.

Answer

Please see attched sheet for solution.

Answer

Answer is D (196 N). Please see the attached sheet for solution.

Answer

Answer is  C. Please see the attached sheet for solution.

Answer

Answer is B (5h/4)

The body will come to rest after an infinite (very very large) number of collisions. We will have to find sum of an infinite series

Let the velocity just before the 1st collision be V0 and the coefficient of restitution be C

V0 = (2gh)^(1/2)

Velocity just after 1st collision = V1 = C*V0

Velocity just after 2nd collision = V2 = C*V1

Velocity just after nth collision = Vn = C*Vn-1

Distance travelled before 1st collision = d0 = h

Distance travelled between 1st and 2nd collision = d1 = 2* (V1^2/2g) = V1^2/g

Distance travelled between 2nd and 3rd collision = d2 = V2^2/g

Distance travelled between nth and n+1th collision = dn = Vn^2/g

Total distance travelled = D = d0 + d1 + d2 +.............+ dn + .............

= h + (1/g) * [C^2*V0^2 + C^4 * V0^2 + ..................... + C^2n * V0^2 +..........

= h + (C^2 * V0^2)/g * [ 1 + C^2 + C^4 +.......................

= h + (C^2 * 2gh)/g * [ 1/(1-C^2)]     ......... (sum of infinite GP)

Putting C=1/3

D = 5h/4