Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:

### Related Formula Springs in parallel: k_p = k_1 + k_2$k_p = k_1 + k_2$ Springs in series: frac1k_s = frac1k_p + frac1k_3$$\frac{1}{k_s} = \frac{1}{k_p} + \frac{1}{k_3}$$ Time period: T = 2pi sqrtfracMk_eq$$T = 2\pi \sqrt{\frac{M}{k_{eq}}}$$ ### Core Logic Evaluate the equivalent spring constant of the given setup by first resolving the parallel configuration, then the series configuration. ### Step 1: Equivalent Spring Constant Based on the standard layout of the problem (as implied by the solution text), let's assume one section is 2k$2k$ and k$k$ in parallel... Wait, the solution implies frac2k cdot k3k + k$\frac{2k \cdot k}{3k} + k$, which represents a series combination of 2k$2k$ and k$k$, followed by a parallel combination with another k$k$. Let's trust the official PDF's algebraic step: k_eq = frac2k cdot k3k + k$k_{eq} = \frac{2k \cdot k}{3k} + k$. This evaluates to: k_eq = frac2k3 + k = frac5k3$$k_{eq} = \frac{2k}{3} + k = \frac{5k}{3}$$ ### Step 2: Time Period Calculation T = 2pi sqrtfracMk_eq$$T = 2\pi \sqrt{\frac{M}{k_{eq}}}$$ T = 2pi sqrtfracM5k/3 = 2pi sqrtfrac3M5k$$T = 2\pi \sqrt{\frac{M}{5k/3}} = 2\pi \sqrt{\frac{3M}{5k}}$$ ### Step 3: Matching the Format Bring the 2 inside the square root as 4: T = pi sqrtfrac4 times 3M5k = pi sqrtfrac12M5k$$T = \pi \sqrt{\frac{4 \times 3M}{5k}} = \pi \sqrt{\frac{12M}{5k}}$$ Comparing with pi sqrtfracalpha M5k$\pi \sqrt{\frac{\alpha M}{5k}}$: alpha = 12$\alpha = 12$ ### Pattern Recognition Whenever an external coefficient like 2 (from 2pi$2\pi$) is pushed inside the square root to match a format, it squares. Don't forget 2 to 4$2 \to 4$, turning 3M$3M$ into 12M$12M$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula v = omega sqrtA^2 - x^2$$v = \omega \sqrt{A^2 - x^2}$$ ### Core Logic At displacement x = frac2A3$x = \frac{2A}{3}$, the initial velocity is: v = omega sqrtA^2 - left(frac2A3right)^2$$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2}$$ v = omega sqrtA^2 - frac4A^29 = omega sqrtfrac5A^29$$v = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}}$$ v = fracsqrt5Aomega3$$v = \frac{\sqrt{5}A\omega}{3}$$ ### Step 2: Velocity Tripled The speed is now increased to v' = 3v$v' = 3v$ at the same position x = frac2A3$x = \frac{2A}{3}$. v' = 3 left(fracsqrt5Aomega3right) = sqrt5Aomega$$v' = 3 \left(\frac{\sqrt{5}A\omega}{3}\right) = \sqrt{5}A\omega$$ This new velocity corresponds to a new amplitude A'$A'$: v' = omega sqrt(A')^2 - x^2$$v' = \omega \sqrt{(A')^2 - x^2}$$ sqrt5Aomega = omega sqrt(A')^2 - left(frac2A3right)^2$$\sqrt{5}A\omega = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2}$$ ### Step 3: Finding new Amplitude Squaring both sides: 5A^2 = (A')^2 - frac4A^29$$5A^2 = (A')^2 - \frac{4A^2}{9}$$ (A')^2 = 5A^2 + frac4A^29 = frac45A^2 + 4A^29$$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9}$$ (A')^2 = frac49A^29$$(A')^2 = \frac{49A^2}{9}$$ A' = frac7A3$$A' = \frac{7A}{3}$$ Comparing with fracnA3$\frac{nA}{3}$, we get n = 7$n = 7$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations