The given function is: \[ f(x) = \sin(x^2). \]
Differentiate \( f(x) \) w.r.t. \( x \): \[ \frac{d}{dx} [\sin(x^2)] = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x. \] At \( x = \sqrt{\pi} \): \[ \frac{d}{dx} [\sin(x^2)] = 2x \cos(x^2) \quad \text{and substitute } x = \sqrt{\pi}: \] \[ \frac{d}{dx} [\sin(x^2)] = 2\sqrt{\pi} \cos(\pi). \] Since \( \cos(\pi) = -1 \): \[ \frac{d}{dx} [\sin(x^2)] = 2\sqrt{\pi} \cdot (-1) = -2\sqrt{\pi}. \]
Therefore, the correct answer is (C) \( -2\sqrt{\pi} \).
Balance Sheet of Chandan, Deepak and Elvish as at 31st March, 2024
Liabilities | Amount (₹) | Assets | Amount (₹) |
---|---|---|---|
Capitals: | Fixed Assets | 27,00,000 | |
Chandan | 7,00,000 | Stock | 3,00,000 |
Deepak | 5,00,000 | Debtors | 2,00,000 |
Elvish | 3,00,000 | Cash | 1,00,000 |
General Reserve | 4,50,000 | ||
Creditors | 13,50,000 | ||
Total | 33,00,000 | Total | 33,00,000 |
A coil of 60 turns and area \( 1.5 \times 10^{-3} \, \text{m}^2 \) carrying a current of 2 A lies in a vertical plane. It experiences a torque of 0.12 Nm when placed in a uniform horizontal magnetic field. The torque acting on the coil changes to 0.05 Nm after the coil is rotated about its diameter by 90°. Find the magnitude of the magnetic field.