Question

A particle is projected at an angle of \(60^\circ\) with the ground. When the projectile makes an angle \(45^\circ\) with the horizontal, its speed becomes \(20\,\text{m s}^{-1}\). Then the initial velocity is:

• A. \(20\sqrt{2}\,\text{m s}^{-1}\)
• B. \(10\sqrt{2}\,\text{m s}^{-1}\)
• C. \(5\sqrt{5}\,\text{m s}^{-1}\)
• D. \(10\sqrt{5}\,\text{m s}^{-1}\)

Hint:

In projectile motion:
Horizontal velocity remains constant
Angle of velocity gives ratio of vertical and horizontal components

Answer 1

The Correct Option is A

Step 1: Determine Standard Cell Potential. Anode (Oxidation): Sn species. \(E^\circ_{ox} = +0.90\) V (Reverse of reduction potential). Cathode (Reduction): Bi species. \(E^\circ_{red} = -0.44\) V. \[ E^\circ_{cell} = E^\circ_{red} + E^\circ_{ox} = -0.44 + 0.90 = 0.46\,\text{V} \]
Step 2: Apply Nernst Equation. \[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \] Determine 'n': Sn goes from +2 to +4 (2e). Bi goes from +3 to 0 (3e). Balanced equation requires transfer of 6 electrons (\(n=6\)). Given \(Q = 10^6\). \[ E_{cell} = 0.46 - \frac{0.06}{6} \log(10^6) \] \[ E_{cell} = 0.46 - (0.01)(6) = 0.46 - 0.06 = 0.40\,\text{V} \]
Step 3: Solve for x. \(E_{cell} = x \times 10^{-1} = 0.4\). \(x = 4\).