2.55 * 105 Ωm
2.55 * 103 Ωm
3.93 * 10-6 Ωm
3.93 * 10-8 Ωm
Correct answer is D
\( R = \frac{1}{a} p \implies p = \frac{R.a}{1} = \frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2 \\
\text{Therefore } p = \frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4} \\
= 0.3925 \times 10^{-7} \text{ OR } -3.930 \times 10 ^{-8}Ωm \)
0.8j
1.8j
9.0j
18.0j
Correct answer is B
Workdone = product of the charge Q, and the p.d V.
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j
The difference observed in solids, liquids and gases can be accounted for by
The different molecules in each of them
The spaces and forces acting between the molecules
Their melting points
Their relative masses
Correct answer is B
No explanation has been provided for this answer.
Which of the following gas laws is equivalent to the work done?
Boyle's law
Charles law
Pressure law
van der Waal's law
Correct answer is A
Mathematically Boyle's law = PV = K
Where P = Pressure
and V = Volume
By definition Pressure = \( \frac{F}{A} = \frac{F}{M \times M} \)
Again Workdone = force x displacement
therefore \( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \\
PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x \)
= force x displacement
4.05m
4.08m
5.05m
6.08m
Correct answer is B
from hooke's law F=Ke
\(
\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}
\frac{4}{0.02} = \frac{15}{e_2} \)
e1 = \( \frac{0.02 x 15}{4} = 0.075M \)
The new length of the string is 4 + 0.075
= 4.075
= 4.08