\(\frac{2}{9}\)ºC
\(\frac{4}{9}\)ºC
\(\frac{9}{2}\)ºC
\(\frac{9}{4}\)ºC
Correct answer is B
Using the law of conservation of energy
mc\(\bigtriangleup\theta\) = mgh
\(\bigtriangleup\theta\) = \(\frac{gh}{c} = \frac{10 \times 20}{450} = \frac{4}{9}\)ºC
\(\frac{p_{0}} {5}\)
\(\frac{4p_{0}}{5}\)
p\(_{0}\)
5P\(_{0}\)
Correct answer is D
p\(_{1}\) = P\(_{0}\), V\(_{1}\) = V\(_{0}\), V\(_{2}\) = \(\frac{1}{5}\)V\(_{0}\) P\(_{2}\) = ?
P\(_{1}\)V\(_{1}\) = P\(_{2}\)V\(_{2}\)
P\(_{2}\) = \(\frac{P_{1}V_{1}}{V_{1}}\)
= \(\frac{\not P_{o}V_{o}}{\frac{1}{5}P_{o}}\) = 5\(\not P_{o}\)
When the temperature of a liquid increases, its surface tension
decreases
increases
remains constant
increases then decreases
Correct answer is A
Surface tension decreases by:
* Increasing liquid's temperature
* Adding soap to the liquid
\(\frac{5}{3}\)
\(\frac{4}{3}\)
\(\frac{3}{4}\)
\(\frac{7}{10}\)
Correct answer is C
R.D = \(\frac{\text {weigh in air - weigh in liquid X}}{\text {weigh in air - weigh in water}}\)
= \(\frac{10 - 7}{10 - 6} = \frac{3}{4}\)
5.0 x 10\(^{-4}\)Nm\(^{-2}\)
5.0 x 10\(^{-5}\)Nm\(^{-2}\)
2.0 x 10\(^{-8}\)Nm\(^{-2}\)
2.0 x 10\(^{9}\)Nm\(^{-2}\)
Correct answer is D
Young's modulus \(\frac{stress}{strain}\)
Strain = \(\frac{e}{l} = \frac{0.05}{10} = \frac{10^{7}}{0.005}\)
2 x 10\(^{9}\)Nm\(^{-2}\)