11.

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If u (O) = 0,  the graph of u(x) versus x will be (where u is potential energy function)

• 

• 

• 

• 

Two blocks of equal masses m are released from the top of a smooth fixed wedge as shown in the figure. The acceleration of the centre of mass of the two blocks is

• g

• $\frac{\mathrm{g}}{2}$

• $\frac{3\mathrm{g}}{4}$

• $\frac{\mathrm{g}}{\sqrt{2}}$

The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge ofthe disc and normal to the disc is

• MR²

• $\frac{1}{2}{\mathrm{MR}}^2$

• $\frac{3}{2}{\mathrm{MR}}^2$

• $\frac{7}{2}{\mathrm{MR}}^2$

Two spherical bodies of mass M and 5 M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller bodyjust before collision, is

• 1.5 R

• 2.5 R

• 4.5 R

• 7.5  R

A satellite orbiting the earth in a circular orbit of radius R completes one revolution in 3 h. If orbital radius of geostationarysatellite is 36,000 km, orbital radius of earth is 

• 6000 km

• 9000 km

• 12000 km

• 15000 km

If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

• increase

• remain unchanged

• decrease

• first increase then decrease

The kinetic energy of an electron, which is accelerated in the potential difference of l00 V, is

• 1.6 × 10^-17 J

• 1.6 × 10^-14 J

• 1.6 × 10^-10 J

• 1.6 × 10^-8 J

A wire canbe brokenbyapplyinga load of200 N. The force required to break another wire of the same length and same material, but double in diameter, is

•  200 N

• 400 N

• 600 N

• 800 N

A cube of side 40 mm has its upper face displaced by 0.1 mm by a  tangenbtial force of 8 kN. The shearing modulus of cube is

• 2 × 10⁹ Nm^-2

• 4 $$\begin{gathered} \mathrm{Shearing} \mathrm{modulus} \mathrm{of} \mathrm{cube} \\ \mathrm{\eta } = \frac{\mathrm{Fl}}{\mathrm{A}∆\mathrm{l}} = \frac{8 \times {10}^3 \times 40 \times {10}^{-3}}{{\left(40 \times {10}^{-3}\right)}^2 \times \left(0.1 \times {10}^{-3}\right)} \\ = 2 \times {10}^{9 } {\mathrm{Nm}}^{-2} \end{gathered}$$ Nm^-2

• 8 × 10⁹ Nm^-2

• 16 × 10⁹ Nm^-2

The neck and bottom of a bottle are3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is

• 30 N

• 150 N

• 300 N

• 600 N