Two springs A and B have force constants $k_{1}$ and $k_{2}$ respectively. The ratio of work done on A to that done on B when they are stretched by the same force is

Q. For a reversible reaction

A_{1} B

the initial molar concentration of

A

and

B

are

^{'} a^{'}

and

^{'} b^{'}

M respectively. If

^{'} x^{'}

mol/litre of

A

is reacted till the achievement of equilibrium, then

^{'} x^{'}

is:

(

K_{1}

and

K_{2}

are rate constants)

Two slabs A and B of different materials but of the same thickness are joined as shown in figure. The thermal conductivities of A and B are $k_{1} a n d k_{2}$ respectively. The thermal conductivity of the composite slab will be

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In a reversible reaction AK1⇌K2B, the initial concentration of A and B are a and b in moles per Litre and the equilibrium concentration are (a - x) and (b + x) respectively ; express x in terms of K1,K2, a and b :

The correct option is A k1a−k2bk1+k2KC=K1K2=b+xa−xSolving the equation for x, we getx=K1a−K2bK1+K2Hence, option A is correct

In a reversible reaction $A_{1} B$ , the initial concentration of A and B are a and b in moles per Litre and the equilibrium concentration are (a - x) and (b + x) respectively ; express x in terms of $K_{1}, K_{2}$ , a and b :

The correct option is A $\frac{k_{1} a - k_{2} b}{k_{1} + k_{2}}$
$K_{C} = \frac{K_{1}}{K_{2}} = \frac{b + x}{a - x}$

Solving the equation for x, we get

$x = \frac{K_{1} a - K_{2} b}{K_{1} + K_{2}}$

Hence, option A is correct