The inner diameter of a brass ring at $273 K$ is $5 c m$ .To what temperature should it be heated for it to accommodate a ball $5) 01 c m$ in diameter. $(α = 2 \times 10^{- 5} /^{o} C)$

273 K

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373 K

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437 K

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173 K

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Solution

The correct option is A $373 K$
In this case, $d_{1} = 5$ cm, $d_{2} = 5.01$ cm
so, $d_{2} - d_{1} = 0.01$ cm
dt $=$ change in temperature $=$ $d l / (α l))$ $=$ 100
Initial temperature $=$ 273 K
So, final temperature $=$ 100 + 273 $=$ 373K

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Q. An iron ring measuring 15.00 cm in diameter is to be shrink on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature

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C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to room temperature.

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Q. An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature

20^{\circ}

C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to room temperature.

α

for iron =

12 \times 10^{- 6} /^{\circ}

Q. An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature

20^{\circ}

C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to room temperature.

α

for iron =

12 \times 10^{- 6} /^{\circ}

An iron ring measuring $15.00 c m$ in diameter is to be shrunk on a pulley which is $15.05 c m$ in diameter. All measurements are done at a room temperature of $20^{\circ} C$ . To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to the room temperature. Coefficient of linear expansion of iron = $12 \times 10^{- 6} /^{\circ} C$ .

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The inner diameter of a brass ring at 273 K is 5 cm.To what temperature should it be heated for it to accommodate a ball 5)01 cm in diameter. (α=2×10−5/oC)

The correct option is A 373 KIn this case, d1=5 cm, d2=5.01 cm so, d2−d1=0.01 cmdt = change in temperature = dl/(αl)) = 100Initial temperature = 273 KSo, final temperature = 100 + 273 = 373K

The inner diameter of a brass ring at $273 K$ is $5 c m$ .To what temperature should it be heated for it to accommodate a ball $5) 01 c m$ in diameter. $(α = 2 \times 10^{- 5} /^{o} C)$

The correct option is A $373 K$
In this case, $d_{1} = 5$ cm, $d_{2} = 5.01$ cm
so, $d_{2} - d_{1} = 0.01$ cm
dt $=$ change in temperature $=$ $d l / (α l))$ $=$ 100
Initial temperature $=$ 273 K
So, final temperature $=$ 100 + 273 $=$ 373K