We are given a thermodynamic process involving temperature changes. Let’s break down the problem step by step:
We start with the following temperature relationships:
We are given the temperature changes over a time interval and can use the following relationship between the initial and final temperatures:
\(T_0 = T\)
The first equation is based on the temperature change from \( 3T \) to \( 2T \) over 10 minutes:
\(\left(\frac{3T - 2T}{10}\right) = c_1 \left(\frac{3T + 2T}{2} - T\right) \) ....(\)
This equation is based on the fact that heat exchange rate is proportional to the temperature difference.
The second equation is based on the temperature change from \( 2T \) to \( T_f \) over 10 minutes:
\(\left(\frac{2T - T_f}{10}\right) = c_1 \left(\frac{2T + T_f}{2} - T\right) \) ....(i\)
Next, we need to take the ratio of the two equations (i) and (ii) to solve for the final temperature \( T_f \):
\(\frac{E(i)}{E(ii)} \Rightarrow \frac{\frac{T}{10}}{\frac{2T - T_f}{10}} = \frac{\frac{5T - 2T}{2}}{\frac{T_f}{2}}\)
After simplifying the equation, we get:
\(\frac{T}{2T - T_f} = \frac{3T}{T_f}\)
Rearranging this equation to solve for \( T_f \), we get:
\(T_f = 6T - 3T_f\)
Now, simplify further:
\(4T_f = 6T\)
Finally, we find the value of \( T_f \):
\(T_f = \frac{3}{2} T\)
The final temperature \( T_f \) is:
\(T_f = \frac{3}{2} T\)
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :