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A thin copper wire of length l metre increases in length by 2% when heated through 10ºC. What is the percentage increase in area when a square copper sheet of length l metre is heated through 10ºC

16%

None of the above

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Explanation

Since percentage increase in length = 2 %

Hence, percentage increase in area of square sheet

$= 2 \times 2 %$ = 4%

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A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation a = $b^{α} c^{β} / d^{γ} e^{δ}$ . If the maximum errors in the measurement of b, c, d and e are b₁%, c₁%, d₁% and e₁%, then the maximum error in the value of a determined by the experiment is

( $b_{1} + c_{1} + d_{1} + e_{1}$ )%

( $b_{1} + c_{1} - d_{1} - e_{1}$ )%

( $α b_{1} + β c_{1} - γ d_{1} - δ e_{1}$ )%

( $α b_{1} + β c_{1} + γ d_{1} + δ e_{1}$ )%

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Explanation

$a = b^{α} c^{β} / d^{γ} e^{δ}$

So maximum error in a is given by

${(\frac{Δ a}{a} \times 100)}_{\max} = α . \frac{Δ b}{b} \times 100 + β . \frac{Δ c}{c} \times 100$ $+ γ . \frac{Δ d}{d} \times 100 + δ . \frac{Δ e}{e} \times 100$

$= (α b_{1} + β c_{1} + γ d_{1} + δ e_{1}) %$

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The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00 ± 0.05) Newton and weight in water is (4.00 ± 0.05) Newton. Then the relative density along with the maximum permissible percentage error is

5.0 ± 11%

5.0 ± 1%

5.0 ± 6%

1.25 ± 5%

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Explanation

Weight in air $= (5 .00 \pm 0 .05) N$

Weight in water $= (4 .00 \pm 0 .05) N$

Loss of weight in water $= (1 .00 \pm 0 .1) N$

Now relative density $= \frac{weight in air}{weight loss in water}$

i.e. R . D $= \frac{5 .00 \pm 0 .05}{1 .00 \pm 0 .1}$

Now relative density with max permissible error

$= \frac{5 .00}{1 .00} \pm (\frac{0 .05}{5 .00} + \frac{0 .1}{1 .00}) \times 100$ $= 5 .0 \pm (1 + 10) %$ $= 5 .0 \pm 11 %$

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The resistance R = $\frac{V}{i}$ where V= 100 ± 5 volts and i = 10 ± 0.2 amperes. What is the total error in R

5.2%

$\frac{5}{2}$ %

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Explanation

$∴ {(\frac{Δ R}{R} \times 100)}_{\max} = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$

$= \frac{5}{100} \times 100 + \frac{0 .2}{10} \times 100$ $= (5 + 2) %$ = 7%

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The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is

0.1 s

0.11 s

0.01 s

1.0 s

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Explanation

Average value $= \frac{2 .63 + 2 .56 + 2 .42 + 2 .71 + 2 .80}{5}$

$= 2 .62 \sec$

Now $| Δ T_{1} | = 2 .63 - 2 .62 = 0 .01$

$| Δ T_{2} | = 2 .62 - 2 .56 = 0 .06$

$| Δ T_{3} | = 2 .62 - 2 .42 = 0 .20$

$| Δ T_{4} | = 2 .71 - 2 .62 = 0 .09$ **

$| Δ T_{5} | = 2 .80 - 2 .62 = 0 .18$

Mean absolute error

$Δ T = \frac{| Δ T_{1} | + | Δ T_{2} | + | Δ T_{3} | + | Δ T_{4} | + | Δ T_{5} |}{5}$

$= \frac{0 .54}{5} = 0 .108 = 0 .11 s e c$

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The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be

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Explanation

Volume of cylinder $V = π r^{2} l$

Percentage error in volume

$\frac{Δ V}{V} \times 100 = \frac{2 Δ r}{r} \times 100 + \frac{Δ l}{l} \times 100$

$= (2 \times \frac{0 .01}{2 .0} \times 100 + \frac{0 .1}{5 .0} \times 100)$ = (1 + 2)% = 3%

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According to Joule's law of heating, heat produced H = I²Rt, where I is current, R is resistance and t is time. If the errors in the measurement of I, R and t are 3%, 4% and 6% respectively then error in the measurement of H is

± 17%

± 16%

± 19%

± 25%

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Explanation

$H = I^{2} R t$ $∴ \frac{Δ H}{H} \times 100 = (\frac{2 Δ I}{I} + \frac{Δ R}{R} + \frac{Δ t}{t}) \times 100$ $= (2 \times 3 + 4 + 6) %$ = 16%

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A physical quantity P is given by P = $\frac{A^{3} B^{\frac{1}{2}}}{C^{- 4} D^{\frac{3}{2}}}$ . The quantity which brings in the maximum percentage error in P is

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Explanation

Quantity C has maximum power. So it brings maximum error in P.

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If L = 2.331 cm, B = 2.1 cm, then L + B =

4.431 cm

4.43 cm

4.4 cm

4 cm

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Explanation

Given, L = 2.331 cm

= 2.33 (correct upto two decimal places)

and B = 2.1 cm = 2.10 cm

$∴ L + B = 2 .33 + 2 .10 = 4 .43 c m$ = 4.4 cm

Since minimum significant figure is 2.

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The number of significant figures in all the given numbers 25.12, 2009, 4.156 and $1 .217 \times 10^{- 4}$ is

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Explanation

The number of significant figures in all of the given number is 4.

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