We are given that the interior angles are in arithmetic progression (A.P.)
with a common difference of 6° and the largest angle is 219°. The sum of the interior angles of an \( n \)-sided polygon is given by: \[ \frac{n}{2} \left( 2a + (n-1) \times 6 \right) = (n-2) \times 180 \] where \( a \) is the first angle. Simplifying: \[ an + 3n^2 - 3n = (n-2) \times 180 \] Now, using the condition that the largest interior angle is 219°, we have: \[ a + (n-1) \times 6 = 219 \] which simplifies to: \[ a = 225 - 6n \] Substitute this value of \( a \) into the sum equation: \[ (225 - 6n) + 3n^2 - 3n = (n-2) \times 180 \] Solving the resulting quadratic equation gives \( n = 20 \).
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.