Two simple pendulum with heavy bobs-one using iron wire and the other aluminimum wire are excited simultaneously. It is found that when the first pendulum completes 1000 oscillations the other completes 1001. When the temperature is raised by `t^(@)C`, it is found that the two pendulums now oscillate together. If he coefficients of thermal expansion of iron and aluminimum are `10 xx 10^(-6)//^(@)C` and `30 xx 10^(-6)//^(@)C` then the value of t is :

Options

A. `77.2^(@)C`
B. `123.2^(@)C`
C. `100.1^(@)C`
D. `105.2^(@)C`

Explanation

To solve the problem, we need to find the temperature increase \( t \) at which the two pendulums oscillate together after being excited. We know that the time period of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. The time period is directly proportional to the square root of the length of the pendulum.

### Step 1: Establish the relationship between the time periods of the two pendulums.

Let \( T_1 \) be the time period of the pendulum with the iron wire and \( T_2 \) be the time period of the pendulum with the aluminum wire. From the problem, we know:

\[ \frac{T_1}{T_2} = \sqrt{\frac{L_1}{L_2}} \]

Given that when the first pendulum completes 1000 oscillations, the second completes 1001, we can express this as:

\[ \frac{T_1}{T_2} = \frac{1001}{1000} \]

### Step 2: Relate the lengths of the pendulums.

From the previous relationship, we can square both sides:

\[ \frac{L_1}{L_2} = \left(\frac{1001}{1000}\right)^2 \]

Calculating this gives:

\[ \frac{L_1}{L_2} = \frac{1002001}{1000000} \approx 1.002001 \]

Thus, we can approximate:

\[ L_1 \approx 1.002 L_2 \]

### Step 3: Set up the equation for the lengths after temperature increase.

When the temperature is raised by \( t \) degrees Celsius, the lengths of the pendulums change due to thermal expansion. The new lengths can be expressed as:

\[ L_1' = L_1(1 + \alpha_1 t) \]
\[ L_2' = L_2(1 + \alpha_2 t) \]

where \( \alpha_1 = 10 \times 10^{-6} \, \text{°C}^{-1} \) (iron) and \( \alpha_2 = 30 \times 10^{-6} \, \text{°C}^{-1} \) (aluminum).

### Step 4: Set the new lengths equal.

Since the pendulums oscillate together after the temperature increase, we have:

\[ L_1(1 + \alpha_1 t) = L_2(1 + \alpha_2 t) \]

Substituting \( L_1 = 1.002 L_2 \):

\[ 1.002 L_2(1 + \alpha_1 t) = L_2(1 + \alpha_2 t) \]

Dividing both sides by \( L_2 \) (assuming \( L_2 \neq 0 \)):

\[ 1.002(1 + \alpha_1 t) = 1 + \alpha_2 t \]

### Step 5: Substitute the values of \( \alpha_1 \) and \( \alpha_2 \).

Substituting \( \alpha_1 = 10 \times 10^{-6} \) and \( \alpha_2 = 30 \times 10^{-6} \):

\[ 1.002(1 + 10 \times 10^{-6} t) = 1 + 30 \times 10^{-6} t \]

Expanding the left side:

\[ 1.002 + 1.002 \times 10 \times 10^{-6} t = 1 + 30 \times 10^{-6} t \]

### Step 6: Rearranging the equation.

Rearranging gives:

\[ 1.002 + 1.002 \times 10^{-5} t = 1 + 30 \times 10^{-6} t \]

Subtract \( 1 \) from both sides:

\[ 0.002 + 1.002 \times 10^{-5} t = 30 \times 10^{-6} t \]

### Step 7: Combine like terms.

Rearranging gives:

\[ 0.002 = (30 \times 10^{-6} t - 1.002 \times 10^{-5} t) \]

This simplifies to:

\[ 0.002 = (30 - 10.02) \times 10^{-6} t \]

\[ 0.002 = 19.98 \times 10^{-6} t \]

### Step 8: Solve for \( t \).

Now, solving for \( t \):

\[ t = \frac{0.002}{19.98 \times 10^{-6}} \approx 100.1 \, \text{°C} \]

### Final Answer:

The value of \( t \) is approximately \( 100.1 \, \text{°C} \).

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