[" e The densities of sakium (Ca) and harium (Ba) are l ss and 35"1" pes "],[" respectively,Based on Dobereinct's lavi of triacls can yog fine "],[" approwimate density of strontium (Sr)? "]

The densities of calcium (ca ) and barium (Ba ) are 1.55 and 3.51 g cm ^(-3) respectively based on Dobereiner's law of triads , can you give the approximate density of strotium (Sr ) ?

Value of absolute zero of temperature in degree Celsius (.^(@)C) can be determined by given data. The density of an ideal gas at 25^(@)C and 100^(@)C are 1.5 and 1.2g/L, respectively, both at the same pressure. The value of absolute zero of temperature in degree Celsius (.^(@)C) :

A small retangular coil ABCD contains 140 turns of wire. The sides AB and BC of the coil are of length 4.5 and 2.8 cm respectively, as shown in the figure The coil is held between the poles of a large magnet so that the coil can rotate about an axis thorugh its centre. The magnet produces a uniform magnetic field of flux density B between its poles. When the current in the coil is 170 mA, the maximum torque produced in the coil is 2.1 xx 10^-3 N m . The current in the coil in (a) is switched off and the coil s positioned as shown in the figure. The coil is then turned thorugh an angle of 90^@ in a time of 0.14 s. Calculate the average e.m.f. induced in the coil.

A cylindrical container of height 3L and cross sectional area A is fitted with a smooth movable piston of negligible weight. It contains an ideal diatomic gas. Under normal atmospheric pressure P_(0) the piston stays in equilibrium at a height L above the base of the container. The gas chamber is provided with a heater and a copper coil through which a cold liquid can be circulated to extract heat from the gas. Volume occupied by the heater and the liquid coil is negligible. Following set of operations are performed to take the gas through a cyclic process. (1) Heater is switched on. At the same time a tap above the cylinder is opened. Water fills slowly in the container above the piston and it is observed that the piston does not move. Water is allowed to fill the container so that the height of water column becomes L. Now the tap is closed. (2) The heater is kept on and the piston slowly moves up. Heater is switched off at the time water is at brink of overflowing. (3) Now the cold liquid is allowed to pass through the coil. The liquid extracts heat from the gas. Water is removed from the container so as to keep the position of piston fixed. Entire water is removed and the gas is brought back to atmospheric pressure. (4) The circulation of cold liquid is continued and the piston slowly falls down to the original height L above the base of the container. Circulation of liquid is stopped. Assume that the container is made of adiabatic wall and density of water is rho . Force on piston due to impact of falling water may be neglected. (a) Draw the entire cycle on a P–V graph. (b) Find the amount of heat supplied by the heater and the amount of heat extracted by the cold liquid from the gas during the complete cycle.

When a jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F=(rhoAV_(0))V_(0)-(rhoAV_(0))V_(0)costheta=rhoAV_(0)^(2)[1-costheta] If surface is free and starts moving due to thrust of the liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let, at any instant, the velocity of surface be u . Then the above equation becomes F=rhoA(V_(0)-u)^(2)[1-costheta] Based on the above concept, as shown in the following figure, if the cart is frictionless and free to move in the horizontal direction, then answer the following. Given that cross sectional area of jet =2xx10^(-4)m^(2) velocity of jet V_(0)=10m//s density of liquid =1000kg//m^(3) ,mass of cart M=10 kg . Initially ( l=0 ) the force on the cart is equal to

When a jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F=(rhoAV_(0))V_(0)-(rhoAV_(0))V_(0)costheta=rhoAV_(0)^(2)[1-costheta] If surface is free and starts moving due to thrust of the liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let, at any instant, the velocity of surface be u . Then the above equation becomes F=rhoA(V_(0)-u)^(2)[1-costheta] Based on the above concept, as shown in the following figure, if the cart is frictionless and free to move in the horizontal direction, then answer the following. Given that cross sectional area of jet =2xx10^(-4)m^(2) velocity of jet V_(0)=10m//s density of liquid =1000kg//m^(3) ,mass of cart M=10 kg . Initially ( l=0 ) the force on the cart is equal to