
(Considering that all isomers satisfy the given conditions) Since given compound produces pink colour with neutral FeCl3. So, it must have at least one –OH group attached to benzene ring.
Also, given compound is optically active, it means it should have a chiral centre, which is only possible when it has –CH(OH)CH3.
DBE (Double bond equivalent) of C8H10O2 is
= \(\text{Number of carbon atoms} - \left(\frac{\text{Number of monovalent atoms}}{2}\right) + 1\)
= \(8 - \frac{10}{2} + 1 = 4\)
So, total isomeric structures = 6
Young double slit arrangement is placed in a liquid medium of 1.2 refractive index. Distance between the slits and screen is 2.4 m.
Slit separation is 1 mm. The wavelength of incident light is 5893 Å. The fringe width is:
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 