Describe how a man weighing 980 N might be able to slide down a rope that can support a weight of only 755 N.

With what minimum acceleration can a fireman slide down a rope whose breaking strength is 2/3 rd of his weight?

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?

If a man weighs 250 N on the surface of earth, what will his wight half way down to the centre of earth which is sipposed to be spherical of uniform mass density.

A long horizontal rigidly supported wire carries a current i_a of 100 A. Directly above it and parallel to its is a fine wire that carries a current i_b of 20 A and weighs 0.073 N/m. How far above the lower wire should the second wire be kept if we wish to support it by magnetic repulsion? Given permeability constant mu_0= 4pi xx 10^-7 Wb A^-1m^-1 .

A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for at least 5 hours a day. She produces only two items M and N each on the three machines are given in the following table: SHe makes a profit of Rs 600 and Rs 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum Profit?

A monkey of mass 40kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. in which of the following cases will the rope break: the monkey:- climbs down with an acceleration of 4 m s^-2

An object weighs 10 N in air. When immessed fully in liquid, it weighs only 8 N. The weight of the liquid displaced by the object will be:

Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =initial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Half-life period of U is 2.5xx10^(5) years. In how much time will the amount of U^(237) remaining be only 25% of the original amount ? a) 2.5xx10^(5) year b) 1.25xx10^(5) years c) 5xx10^(5) years d) none of these