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IIT-JEE 2011 Paper 1 Offline

Numerical

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.2 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is _________.

Your Input ________

Answer

Correct Answer is 4

Explanation

f = $$\mu$$mg

The net torque about point P is

F $$\times$$ R $$-$$ fR = Ip$$\alpha$$

Where, Ip = mR2 + mR2 = 2mR2 (parallel axes theorem)

and, a = R$$\alpha$$

Also, f = $$\mu$$mg

$$\therefore$$ $$F \times R - \mu mgR = (2\,m{R^2}) \times \left( {{a \over R}} \right) = 2maR$$

$$ \Rightarrow F - \mu mg = 2\,ma$$

$$ \Rightarrow 2 - \mu \times 2 \times 10 = 2 \times 2 \times 0.3$$

which gives $$\mu = {{0.8} \over {2 \times 10}} = {{0.4} \over {10}}$$. Hence, P = 4

2

IIT-JEE 2011 Paper 1 Offline

Numerical

Four solid spheres each of diameter $$\sqrt 5 $$ cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N $$ \times $$ 10−4 kg-m2, then N is

Your Input ________

Answer

Correct Answer is 9

Explanation

The moment of inertia of each sphere about an axis passing through its centre is $${2 \over 5}m{r^2}$$.

The moment of inertia of sphere B and sphere D about X $$-$$ X' is

$${I_B} = {I_D} = {2 \over 5}m{r^2}$$.

Using parallel axes theorem, the moment of inertia of sphere A and sphere C about X $$-$$ X' is

$${I_A} = {I_C} = {2 \over 5}m{r^2} + m{d^2}$$

The moment of inertia of the system about the diagonal is

$$I = {I_A} + {I_B} + {I_C} + {I_D} = {8 \over 5}m{r^2} + 2m{d^2}$$

$$m = 0.5$$ kg

$$d = {a \over {\sqrt 2 }} = {4 \over {\sqrt 2 }}$$ cm

$$r = {{\sqrt 5 } \over 2}$$ cm

$$\therefore$$ $$I = {8 \over 5} \times 0.5 \times {\left( {{{\sqrt 5 } \over 2} \times {{10}^{ - 2}}} \right)^2} + 2 \times 0.5 \times {\left( {{4 \over {\sqrt 2 }} \times {{10}^{ - 2}}} \right)^2}$$

$$ = 9 \times {10^{ - 4}}$$ kg m2

Hence, $$N = 9$$

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