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A stone in free fall covers distances $h_1$, $h_2$, $h_3$ in successive 5-second intervals. The correct relation is:
Distances in successive equal time intervals from rest are in the ratio of odd numbers: $1:3:5:7\ldots$
So $h_1 : h_2 : h_3 = 1 : 3 : 5$, i.e., $\mathbf{h_1 = \dfrac{h_2}{3} = \dfrac{h_3}{5}}$
Entropy is a state function. This means the change in entropy depends only on the initial and final states of the system, not on the path or the number of intermediate steps taken to reach that state.
Since the body is heated from an initial temperature $T_1 = 373 \text{ K}$ to a final temperature $T_2 = 473 \text{ K}$ in both cases, the entropy change $\Delta S$ of the body will be identical in both scenarios.
$\Delta S = \int \frac{dQ}{T} = \int_{T_1}^{T_2} \frac{ms dT}{T} = C \ln \left( \frac{T_2}{T_1} \right)$.
$S_n = \dfrac{q^{n+1}-1}{q-1}$. The LHS sum involves $\sum_{k=1}^{101}\binom{101}{k}S_{k-1}$.
After substituting and applying binomial theorem: the sum simplifies using $\sum \binom{101}{k}\frac{q^k-1}{q-1}$.
The result evaluates to $\alpha T_{100}$ where $\alpha = 2^{100}$.
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