SATHEE: Physics And Measurement Question 176

Physics And Measurement Question 176

Question: 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 $\pm 0.05$ ) Newton and weight in water is (4.00 $\pm$ 0.05) Newton. Then the relative density along with the maximum permissible percentage error is

Options:

A) 5.0 $\pm$ 11%

B) 5.0 $\pm$ 1%

C) 5.0 $\pm$ 6%

D) 1.25 $\pm$ 5%

Show Answer

Answer:

Correct Answer: A

Solution:

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 $= \frac{w e i g h t i n air}{weight loss in water}$

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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Physics And Measurement Question 176

Question: 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

Options:

Answer:

Solution:

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Question: 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 $\pm 0.05$ ) Newton and weight in water is (4.00 $\pm$ 0.05) Newton. Then the relative density along with the maximum permissible percentage error is