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NVAEasyJEE 2025Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2025 Question with Solution

The increase in pressure required to decrease the volume of a water sample by 0.2%0.2\% is P×105 N m−2P \times 10^5 \, \text{N m}^{-2}. Bulk modulus of water is 2.15×109 N m−22.15 \times 10^9 \, \text{N m}^{-2}. The value of PP is _____.

Answer

Correct answer:43

Step-by-step solution

Standard Method

Given: The decrease in volume is 0.2%0.2\%, so

ΔVV=−0.2100=−0.002\frac{\Delta V}{V} = -\frac{0.2}{100} = -0.002

The bulk modulus of water is

B=2.15×109 N m−2B = 2.15 \times 10^9 \, \text{N m}^{-2}

The required pressure increase is written as P×105 N m−2P \times 10^5 \, \text{N m}^{-2}.

Find: The value of PP.

Using the bulk modulus relation,

B=−ΔPΔV/VB = -\frac{\Delta P}{\Delta V/V}

Substitute the given values:

2.15×109=−ΔP−0.0022.15 \times 10^9 = -\frac{\Delta P}{-0.002}

So,

ΔP=(2.15×109)(0.002)=4.3×106 N m−2\Delta P = (2.15 \times 10^9)(0.002) = 4.3 \times 10^6 \, \text{N m}^{-2}

Now compare with the given form:

P×105=4.3×106P \times 10^5 = 4.3 \times 10^6

Therefore,

P=4.3×106105=43P = \frac{4.3 \times 10^6}{10^5} = 43

Hence, the value of PP is 4343.

Direct Substitution

Given: B=2.15×109 N m−2B = 2.15 \times 10^9 \, \text{N m}^{-2} and volume decrease =0.2%=0.002= 0.2\% = 0.002.

Find: PP in P×105 N m−2P \times 10^5 \, \text{N m}^{-2}.

Use the magnitude form directly:

ΔP=B∣ΔVV∣\Delta P = B \left|\frac{\Delta V}{V}\right|

So,

ΔP=2.15×109×0.002=4.3×106 N m−2\Delta P = 2.15 \times 10^9 \times 0.002 = 4.3 \times 10^6 \, \text{N m}^{-2}

Therefore,

P=4.3×106105=43P = \frac{4.3 \times 10^6}{10^5} = 43

So the required numerical value is 4343.

Common mistakes

  • Using 0.20.2 instead of 0.0020.002 for 0.2%0.2\%. This is wrong because percentage must be divided by 100100. Always convert 0.2%0.2\% to 0.0020.002 before substitution.

  • Ignoring that volume decreases and mishandling the negative sign in ΔV/V\Delta V/V. This can give a negative pressure change, whereas the question asks for the required increase in pressure. Use the sign convention carefully and report the magnitude of the pressure increase.

  • Stopping at ΔP=4.3×106 N m−2\Delta P = 4.3 \times 10^6 \, \text{N m}^{-2} and not converting it into the form P×105 N m−2P \times 10^5 \, \text{N m}^{-2}. The asked quantity is PP, so divide by 10510^5 to get 4343.

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