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A brass rod and a lead rod each 80 cm lo...

A brass rod and a lead rod each 80 cm long at `0^@C` are clamped together at one end with their free ends coinciding. The separatioin of free ends of the rods if the system is placed in a steam bath is (`alpha_("brass")=18xx10^(-6)//^(@)C` and `alpha_("lead")=28xx10^(-6)//^(@)C`)

A

0.2 mm

B

0.8 mm

C

1.4 mm

D

1.6 mm

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The correct Answer is:
To solve the problem of finding the separation of the free ends of a brass rod and a lead rod when placed in a steam bath, we will follow these steps: ### Step 1: Understand the Problem We have two rods, one made of brass and the other made of lead, both initially 80 cm long at 0°C. They are clamped at one end, and we need to find the separation of their free ends when the system is heated to 100°C (the temperature of the steam bath). ### Step 2: Identify the Coefficients of Linear Expansion The coefficients of linear expansion for the materials are given as: - For brass, \( \alpha_{brass} = 18 \times 10^{-6} \, /^{\circ}C \) - For lead, \( \alpha_{lead} = 28 \times 10^{-6} \, /^{\circ}C \) ### Step 3: Calculate the Change in Length for Each Rod We will use the formula for linear expansion: \[ L' = L + \alpha \cdot L \cdot \Delta T \] where: - \( L' \) is the final length, - \( L \) is the initial length, - \( \alpha \) is the coefficient of linear expansion, - \( \Delta T \) is the change in temperature. The initial length \( L \) for both rods is 80 cm, and the change in temperature \( \Delta T \) is \( 100°C - 0°C = 100°C \). #### For the Brass Rod: \[ L'_{brass} = 80 \, \text{cm} + (18 \times 10^{-6}) \cdot 80 \cdot 100 \] Calculating: \[ L'_{brass} = 80 + 0.144 \, \text{cm} = 80.144 \, \text{cm} \] #### For the Lead Rod: \[ L'_{lead} = 80 \, \text{cm} + (28 \times 10^{-6}) \cdot 80 \cdot 100 \] Calculating: \[ L'_{lead} = 80 + 0.224 \, \text{cm} = 80.224 \, \text{cm} \] ### Step 4: Calculate the Separation of the Free Ends The separation of the free ends of the rods is given by the difference in their final lengths: \[ \text{Separation} = L'_{lead} - L'_{brass} \] Substituting the values we calculated: \[ \text{Separation} = 80.224 \, \text{cm} - 80.144 \, \text{cm} = 0.08 \, \text{cm} \] ### Step 5: Convert to Meters (if needed) To express the separation in meters: \[ 0.08 \, \text{cm} = 0.0008 \, \text{m} = 8 \times 10^{-4} \, \text{m} \] ### Final Answer The separation of the free ends of the rods when placed in a steam bath is \( 0.08 \, \text{cm} \) or \( 8 \times 10^{-4} \, \text{m} \). ---

To solve the problem of finding the separation of the free ends of a brass rod and a lead rod when placed in a steam bath, we will follow these steps: ### Step 1: Understand the Problem We have two rods, one made of brass and the other made of lead, both initially 80 cm long at 0°C. They are clamped at one end, and we need to find the separation of their free ends when the system is heated to 100°C (the temperature of the steam bath). ### Step 2: Identify the Coefficients of Linear Expansion The coefficients of linear expansion for the materials are given as: - For brass, \( \alpha_{brass} = 18 \times 10^{-6} \, /^{\circ}C \) ...
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A brass rod and a lead rod each 80.00 cm long at 0^@C are clamped together at one end with their free ends coinciding. The separation of the free ends of the rods it the system is placed in steam bath is (coefficients of linear expansions of brass and lead are 2xx10^(-5)//^@C and 1.0 xx 10^(-5)//^@C respectively)

The length of the steel rod which would have the same difference in length with a copper rod of length 24cm at all temperatures. (alpha_("copper") = 18 xx 10^(-6) K^(-1) alpha_("steel") = 12 xx 10^(-6) k^(-1)) is -

An iron wire AB has diameter of 0.6 mm and length 3 m at 0^(@)C . The wire is now stretched between the opposite walls of a brass casing at 0^(@)C . What is the extra tension that will be set up in the wire when the temperature of the system is raised to 40^(@)C ? Givne alpha_("brass")=18xx10^(-6)//K alpha_("iron")=12xx10^(-6)//K Y_("iron")=21xx10^(10)N//m_^(2)

A steel rod is 4000cm in diameter at 30^(@)C A brass ring has an interior diameter of 3.992cm at 30^(@) in order that the ring just slides onto the steel rod the common temperature of the two should be nearly (alpha_(steel)=11xx10^(-6)/(^(@)C) and alpha_(brass)=19xx10^(-6)/(^(@)C)

The brass scale of a barometer gives correct reading at 10^(@)C . The baometer reads 75 cm at 30^(@)C . What is the a atmoshpheric pressure at o^(@)C ("in cm "Hg) (alpha_("brass")=20xx10^(-6)//^(0)C,lambda_(Hg)=175xx10^(-6)//^(@)C)

A heavy brass bar has projections at its ends as shown in the figure. Two find steel wires, fastened between the projections, are just taut (zero tension) when the whole systeam is at 0^(0)C . What is the tensile stress in the steel wires when the temperature of the systeam is raised to 300^(0)C ? Given that alpha_("brass") = 20 xx 10^(-6 ^(@))C^(-1) alpha_("steel") = 12 xx 10^(-6@)C^(-1) Y_("steel") = 2 xx 10^(11) Nm^(-2)

A clock is calibrated at a temperature of 20^@C Assume that the penduum is a thin brass rod of negligible mass with a heavy bob attached to the end (alpha_("brass")=19xx10^(-6)//K)

A steel rod is 3.000cm in diameter at 25^(@)C . A brass ring has an interior diameter of 2.9912cm at 25^(@)C . At what common temperature will the ring just slide onto the rod? alpha_("brass")=18xx10^(-6)K^(-1) and alpha_("steel")=12xx10^(-6)K^(-1)

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