SATHEE: Physics And Measurement Question 197

Physics And Measurement Question 197

Question: The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is $W_{1}^{}$ = $8.00 \pm 0.05 N$ and the weight in water is $W_{2}^{}$ = $6.00 \pm 0.05$ N, then the relative density $ρ_{r}^{} = W_{1}^{} / (W_{1}^{} - W_{2}^{})$ With the maximum permissible error is

Options:

A) $4.00 \pm 0.62$ %

B) $4.00 \pm 0.82$ %

C) $4.00 \pm 3.2$ %

D) $4.00 \pm 5.62$ %

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Relative density $ρ_{r} = \frac{W_{1}}{W_{1} - W_{2}} = \frac{8.00}{8.00 - 6.00} = 4.00$

$\frac{Δ ρ_{r}}{ρ_{r}} \times 100 = \frac{Δ W_{1}}{W_{1}} \times 100 + \frac{Δ (W_{1} - W_{2})}{W_{1} - W_{2}} \times 100$

$= \frac{0.05}{8.00} \times 100 + \frac{0.05 + 0.05}{2} \times 100 = 5.62$ %

$∴ ρ_{r} = 4.00 \pm 5.62$ %

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Physics And Measurement Question 197

Question: The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is W1 = 8.00±0.05N and the weight in water is W2 = 6.00±0.05 N, then the relative density ρr=W1/(W1−W2) With the maximum permissible error is

Options:

Answer:

Solution:

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