Find time taken by the block to slide a distance \(8.8\,\text{m}\) with respect to the wedge (in seconds). All surfaces are smooth. 
Step 1: Equation for whole system Let acceleration of wedge be \(a\). For horizontal motion of system: \[ 24 = 10a + 2(a + a_r\cos37^\circ) \] \[ 24 = 12a + 2a_r\cos37^\circ \] Using \(\cos37^\circ = \frac{4}{5}\) \[ 24 = 12a + \frac{8a_r}{5} \] \[ 24 = 12a + \frac{8a_r}{5} \quad ...(1) \]
Step 2: Acceleration of block relative to wedge Along the incline: \[ mg\sin37^\circ - ma\cos37^\circ = ma_r \] \[ 2(10)\frac{3}{5} - 2a\frac{4}{5} = 2a_r \] \[ 6 - \frac{4a}{5} = a_r \quad ...(2) \] Step 3: Solve equations Substitute (2) into (1): \[ 24 = 12a + \frac{8}{5}\left(6 - \frac{4a}{5}\right) \] Solving, \[ a_r = 4.92\ \text{m/s}^2 \] Step 4: Time taken Relative displacement: \[ s = \frac12 a_r t^2 \] \[ 8.8 = \frac12 (4.92)t^2 \] \[ t^2 = 3.58 \] \[ t = 1.89\ \text{s} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,