Let $$\eta_{1}$$ is the efficiency of an engine at $$T_{1}=447^{\circ} \mathrm{C}$$ and $$\mathrm{T}_{2}=147^{\circ} \mathrm{C}$$ while $$\eta_{2}$$ is the efficiency at $$\mathrm{T}_{1}=947^{\circ} \mathrm{C}$$ and $$\mathrm{T}_{2}=47^{\circ} \mathrm{C}$$ The ratio $$\frac{\eta_{1}}{\eta_{2}}$$ will be :
An object is taken to a height above the surface of earth at a distance $${5 \over 4}$$ R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :
A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms$$-$$1 gets embedded in it, then loss of kinetic energy will be :
A ball is projected from the ground with a speed 15 ms$$-$$1 at an angle $$\theta$$ with horizontal so that its range and maximum height are equal,
then 'tan $$\theta$$' will be equal to :