Properties Of Solids And Liquids Question 563

Question: A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27C it weighs 30 gm. When the temperature of liquid is raised to 42C , the metal piece weighs 30.5 gm, specific gravity of the liquid at 42C is 1.20, then the linear expansion of the metal will be

Options:

A) 3.316×105/C

B) 2.316×105/C

C) 4.316×105/C

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

Loss of weight at 27C is

=4630=16=V1×1.24ρ1×g

Loss of weight at 42C is

=4630.5=15.5=V2×1.24ρ1×g.(ii)

Now dividing (i) by (ii), we get

1615.5=V1V2×1.241.2

But V2V1=1+3α(t2t1)=15.5×1.2416×1.2=1.001042

3α(4227)=0.001042

α=2.316×105/C



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक