At constant temperature, if pressure increases by $1\% $, the percentage decrease of volume is:
A) $1\% $
B) $\dfrac{{100}}{{101}}\% $
C) $\dfrac{1}{{100}}\% $
D) $\dfrac{1}{{101}}\% $

Hint:To solve this we must know the law that gives the relationship between volume and pressure at constant temperature. Boyle's law gives the relationship between volume and pressure at constant temperature. Consider the initial pressure to be $100{\text{ mm}}$ and then solve.

Complete solution:
We know that Boyle’s law states that at constant temperature the volume of an ideal gas is inversely proportional to the pressure of the gas. Thus,
${P_1}{V_1} = {P_2}{V_2}$ …… (1)
Where ${P_1}$ is the initial pressure,
${P_2}$ is the final pressure,
${V_1}$ is the initial volume,
${V_2}$ is the final volume.
Assume that the initial pressure is $100{\text{ mm}}$. Thus,
${P_1} = 100{\text{ mm}}$
We are given that the pressure increases by $1\% $. Thus, the final pressure will be,
${P_2} = 101{\text{ mm}}$
Substitute the values for the initial pressure and final pressure in equation (1). Thus,
$100{\text{ mm}} \times {V_1} = 101{\text{ mm}} \times {V_2}$
${V_2} = \dfrac{{100{\text{ mm}}}}{{101{\text{ mm}}}} \times {V_1}$
The decrease in the volume will be,
Decrease in volume $ = {V_1} - \dfrac{{100{\text{ mm}}}}{{101{\text{ mm}}}} \times {V_1}$
Decrease in volume $ = \dfrac{{100}}{{101}}\% $
Thus, the percentage decrease of volume is $\dfrac{{100}}{{101}}\% $.
Thus, the correct option is (B) all of the above.

Note:We know that Boyle's law states that at constant temperature the volume of an ideal gas is inversely proportional to the pressure of the gas. Thus, as the volume of the gas increases, the pressure decreases or as the volume of the gas decreases, the pressure increases.