Fringe pattern in YDSE is formed due to the phenomenon of

In YDSE fringe pattern is formed due to interference.

The correct Option is "Interference of light"

Rectillinear propagation of light

Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside

Pressure is reduce, so the temperature falls

Remains at the same temperature

The correct Option is "Starts becoming cooler"

May become hotter or cooler depending upon the amount of water vapour present

The unit of thermal conductivity is:

The heat current related to difference of temperature across the length $l$ of a conductor of area A is $\frac{{{\rm{dH}}}}{{{\rm{dt}}}}\, = \,\frac{{{\rm{KA}}}}{l}\Delta {\rm{T}}\,$ (K = coefficient of thermal conductivity) $\therefore \,K = \frac{{\ell \,dH}}{{A\,\,dt\,\,\Delta T}}$ ${\rm{Unit}}\,{\rm{of}}\,{\rm{K}}{\mkern 1mu} {\rm{ = W}}{\mkern 1mu} {{\rm{m}}^{ - 1}}{\mkern 1mu} {{\rm{K}}^{ - {\rm{1}}}}$

The correct Option is "$W\,{m^{ - 1}}\,{K^{ - 1}}$"

${v_1}\, and \,{v_2}$ are the velocities of sound at the same temperature in two monoatomic gases of densities ${\rho _1}$ and ${\rho _2}$ respectively. If ${\rho _1}/{\rho _2} = \frac{1}{4}$ then the ratio of velocities ${v_1}{\mkern 1mu} \,\,and\,\,{\mkern 1mu} {v_2}$ will be

At given temperature and pressure $v \propto \frac{1}{{\sqrt \rho }} \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{\rho _2}}}{{{\rho _1}}}} = \sqrt {\frac{4}{1}} = 2:1$

If $a$ and $b$ are the nearest and farthest distances of a planet form the sun and the planet is revolving in a elliptical orbit, then square of the time period of revolution of planet is directly proportional to

from kepler’s third law, ${T^2} \propto {R^3}$ ${T^2} \propto {[{\rm{Semi}}\,{\rm{major}}\,{\rm{axis}}]^3}$ Length major axis $ = \frac{{a + b}}{2}$ $\begin{array}{l} {T^2} \propto {\left( {\frac{{a + b}}{2}} \right)^3}\\ {T^2} \propto \frac{{{{\left( {a + b} \right)}^3}}}{8}\\ {T^2} \propto {\left( {a + b} \right)^3} \end{array}$

The correct Option is "${\left( {a + b} \right)^3}$"

Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of $2\;m/{s^2}$ in the upward direction in both the blocks. If T and $T'$ be the tensions in the two parts of the string, then $\left( {\;g = 9.8\;m/{s^2}} \right)$

FBD of mass 2 kg FBD of mass 4kg , $TT' - 19.6 = 4...\left( i \right)$ , $T' - 39.2 = 8...\left( {ii} \right)$ From (ii), $T' = 47.2\;N$ And substituting $T'$ in (i), we get $T = 4 + 19.6 + 47.2 \Rightarrow T = 70.8\;N$

The correct Option is "$T = 70.8\;N\,and\,T' = 47.2\;N$"

$T = 58.8\;N\,and\,T' = 47.2\;N$

$T = 70.8\;N\,and\,T' = 58.8\;N$

A cylindrical capacitor has charge $Q$ and length $L.$ If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor

Capacitance of a cylindrical capacitor $ = \frac{{2\pi {\varepsilon _0}L}}{{{\rm{ln\;}}\left( {b/a} \right)}}$ Energy stored in the capacitor $\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}\frac{{{Q^2}\;{\rm{ln\;}}\left( {b - a} \right)}}{{2\pi {\varepsilon _0}L}} = \left[ {{\rm{const}}} \right]\frac{{{Q^2}}}{L}$ If the charge is doubled and length is doubled, $\left[ {const} \right]\frac{{{Q^{'2}}}}{{L'}} = \frac{4}{2}\left( {\frac{{{Q^2}}}{L}} \right) = 2$ times the energy

The correct Option is "Increases two times"

Two spherical conductors A and B of radii $1\;mm$ and 2 $mm$ are separated by a distance of $5\;cm$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

After connecting through a wire ${V_A} = {V_B}$ $ \Rightarrow \frac{{k{Q_A}}}{{{r_A}}} = \frac{{k{Q_B}}}{{{r_B}}} \Rightarrow \frac{{{Q_A}}}{{{Q_B}}} = \frac{{{r_A}}}{{{r_B}}}$ Ratio of electric field $\frac{{{E_A}}}{{{E_B}}} = \frac{{{Q_A}}}{{{Q_B}}} \times {\left( {\frac{{{r_B}}}{{{r_A}}}} \right)^2}\;\;\;\;\left[ {{ \because }E = \frac{{kQ}}{{{r^2}}}} \right]$ $ \Rightarrow \frac{{{E_A}}}{{{E_B}}} = \frac{{{r_A}}}{{{r_B}}} \times {\left( {\frac{{{r_B}}}{{{r_A}}}} \right)^2} = \frac{{{r_B}}}{{{r_A}}} = \frac{2}{1}$

The ratio of speed of sound in nitrogen and helium gas at $300\,\,K$ is

According to Laplace, the speed of sound in a gas is $v = \sqrt {\frac{{\gamma p}}{d}} = \sqrt {\frac{{\gamma RT}}{M}} $ Where R is gas constant, T the temperature and M the molecular weight. For monatomic gas helium, ${\gamma _1} = \frac{5}{3},{M_1} = 4$ For diatomic gas nitrogen, ${\gamma _2} = \frac{7}{5},{M_2} = 28$ $\begin{array}{l} \therefore \frac{{{v_{{N_2}}}}}{{{v_{{H_e}}}}} = \sqrt {\frac{{{\gamma _1}{M_2}}}{{{\gamma _2}{M_1}}}} \\ = \sqrt {\frac{{5/3}}{{7/5}} \times \frac{{28}}{4}} = \sqrt {\frac{{5 \times 5}}{3}} = \frac{5}{{\sqrt 3 }}\\ Hence,\\ \frac{{{v_h}_2}}{{{v_{He}}}} = \frac{{\sqrt 3 }}{5} \end{array}$

The correct Option is "$\frac{{\sqrt 3 }}{5}$"

Why is refractive index in a transparent medium greater than one?

Absolute refractive index is defined as the ratio of speed of light in free space to that in a given medium $ie,\,\,{\rm{\mu }} = \frac{c}{v}$ For a given light, denser is the medium, lesser will be the speed of light and so greater will be the refractive index, $eg$, as ${v_{{\rm{glass}}}}\,\, > \,\,{v_{{\rm{water}}}},\,\,\,{{\rm{\mu }}_{\rm{g}}} < \,\,{{\rm{\mu }}_w}$. As speed of light in vacuum is always greater than speed in a transparent medium, so the refractive index in a transparent medium is greater than one.