A) 10%
B) \[\left( 10+\frac{2730}{{{t}_{1}}} \right)\]%
C) 20%
D) \[(0.1+t_{1}^{-1})\]%
Correct Answer: B
Solution :
| [b]:\[{{V}_{2}}=\left( {{V}_{1}}+\frac{{{V}_{1}}\times 10}{100} \right)=1.1{{V}_{1}}\] |
| Now,\[{{V}_{2}}\frac{{{V}_{1}}}{({{t}_{1}}+273)}=\frac{1.1{{V}_{1}}}{({{t}_{2}}+273)}\] |
| or\[{{t}_{2}}=(1.1{{t}_{1}}+27.3)\] |
| Increase in temperature \[=1.1{{t}_{1}}+27.3-{{t}_{1}}\] |
| \[=(0.1{{t}_{1}}+27.3)\] |
| Percent increase \[=\frac{(0.1{{t}_{1}}+27.3)\times 100}{{{t}_{1}}}\] |
| \[=\left( 10+\frac{2730}{{{t}_{1}}} \right)\]% |
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