In two different experiments, an object of mass $$5\mathrm{kg}$$ moving with a speed of $$25{\mathrm{ms}}^{-1}$$ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :

Two bodies of mass $$1\mathrm{kg}$$ and $$3\mathrm{kg}$$ have position vectors $$\hat{i}+2\hat{j}+\hat{k}$$ and $$-3\hat{i}-2\hat{j}+\hat{k}$$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :

If momentum of a body is increased by 20%, then its kinetic energy increases by

A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :

A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs$$-$$1 and at a speed of 10 ms$$-$$1. Then, the initial acceleration of the block, in ms$$-$$2, will be :

A water drop of radius 1 $$\mu$$m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2 and its density is negligible as compared to that of water 106 gm$$-$$3. Terminal velocity of the water drop is : (Take acceleration due to gravity = 10 ms$$-$$2)

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity. Reason R : Coefficient of viscosity = {{Force} \over {Velocity\,gradient}} Choose the correct answer from the options given below :

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :

The velocity of a small ball of mass 'm' and density d1, when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is d2, then the viscous force acting on the ball, will be :

If p is the density and $$\eta$$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is :

The terminal velocity (vt) of the spherical rain drop depends on the radius (r) of the spherical rain drop as :

Potential energy as a function of r is given by U = {A \over {{r^{10}}}} - {B \over {{r^5}}}, where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be :

The bulk modulus of a liquid is 3 $$\times$$ 1010 Nm$$-$$2. The pressure required to reduce the volume of liquid by 2% is :

An air bubble of negligible weight having radius r rises steadily through a solution of density $$\sigma$$ at speed v. The coefficient of viscosity of the solution is given by :

A drop of liquid of density $$\rho$$ is floating half immersed in a liquid of density $$\sigma$$ and surface tension $$7.5\times 10^{-4}$$ Ncm$$-$$1. The radius of drop in $$\mathrm{cm}$$ will be : (g = 10 ms$$-$$2)

A water drop of radius $$1\mathrm{cm}$$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $$\mathrm{cm}$$, then the gain in surface energy upto first decimal place will be : (Given $$\pi =3.14$$ )

The area of cross section of the rope used to lift a load by a crane is $$2.5\times 10^{-4}{\mathrm{m}}^2$$. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : (take $$g=10m{s}^{-2}$$ )

Two cylindrical vessels of equal cross-sectional area $$16{\mathrm{cm}}^2$$ contain water upto heights $$100\mathrm{cm}$$ and $$150\mathrm{cm}$$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water $$=10^3\mathrm{kg}/{\mathrm{m}}^3$$ and $$\mathrm{g}=10{\mathrm{ms}}^{-2}$$ ] :

A steel wire of length $$3.2\mathrm{m}\left({\mathrm{Y}}_{\mathrm{s}}=2.0\times 10^{11}{\mathrm{Nm}}^{-2}\right)$$ and a copper wire of length $$4.4\mathrm{m}\left({\mathrm{Y}}_{\mathrm{c}}=1.1\times 10^{11}{\mathrm{Nm}}^{-2}\right)$$, both of radius $$1.4\mathrm{mm}$$ are connected end to end. When stretched by a load, the net elongation is found to be $$1.4\mathrm{mm}$$. The load applied, in Newton, will be: $$($$ Given $$\pi =\frac{22}{7}$$)