JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 12)
If Y, K and $$\eta $$ are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.
$$Y = {{9K\eta } \over {3K - \eta }}N/{m^2}$$
$$Y = {{9K\eta } \over {2\eta + 3K}}N/{m^2}$$
$$\eta = {{3YK} \over {9K + Y}}N/{m^2}$$
$$K = {{Y\eta } \over {9\eta - 3Y}}N/{m^2}$$
Explanation
We know that,
$$Y = 3K(1 - 2\sigma )$$
$$ \Rightarrow \sigma = {1 \over 2}\left( {1 - {Y \over {3K}}} \right)$$ ..... (i)
Also, $$Y = 2\eta (1 + \sigma )$$
$$ \Rightarrow \sigma = {Y \over {2\eta }} - 1$$ .... (ii)
On comparing Eqs. (i) and (ii), we get
$$\left( {1 - {Y \over {3K}}} \right){1 \over 2} = {Y \over {2\eta }} - 1$$
On solving, we get
$$K = {{\eta Y} \over {9\eta - 3Y}}$$ N/m2
$$Y = 3K(1 - 2\sigma )$$
$$ \Rightarrow \sigma = {1 \over 2}\left( {1 - {Y \over {3K}}} \right)$$ ..... (i)
Also, $$Y = 2\eta (1 + \sigma )$$
$$ \Rightarrow \sigma = {Y \over {2\eta }} - 1$$ .... (ii)
On comparing Eqs. (i) and (ii), we get
$$\left( {1 - {Y \over {3K}}} \right){1 \over 2} = {Y \over {2\eta }} - 1$$
On solving, we get
$$K = {{\eta Y} \over {9\eta - 3Y}}$$ N/m2
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