During Searle's experiment, zero of the Vernier scale lies between 3.20 $$ \times $$ 10^-2 m and 3.25 $$ \times $$ 10^-2 m of the main scale. The 20^th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 $$ \times $$ 10^-2 m and 3.25 $$ \times $$ 10^-2 m of the main scale but now the 45^th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 $$ \times $$ 10^-7 m². The least count of the Vernier scale is 1.0 $$ \times $$ 10^-5 m. The maximum percentage error in the Young's modulus of the wire is

Let $${E_1}\left( r \right),{E_2}\left( r \right)$$ and $${E_3}\left( r \right)$$ be the respective electric field at a distance $$r$$ from a point charge $$Q,$$ an infinitely long wire with constant linear charge density $$\lambda ,$$ and an infinite plane with uniform surface charge density $$\sigma .$$ If $$E{}_1\left( {{r_0}} \right) = {E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$$ at a given distance $${r_0}.$$ then

A parallel plate capacitor has a dielectric slab of dielectric constant $$K$$ between its plates that covers $$1/3$$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $$C$$ while that of the portion with dielectric in between is $${C_1}.$$ When the capacitor is charged, the plate area covered by the dielectric gets charge $${Q_1}$$ and the rest of the area gets charge $${Q_2}.$$ The electric field in the dielectric is $${E_1}$$ and that in the other portion is $${E_2}.$$ Choose the correct option/ options, ignoring edge effects.

Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) Joules. The value of n is (take acceleration due to gravity = 10 ms^–2)