If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :

If momentum [P], area $$[\mathrm{A}]$$ and time $$[\mathrm{T}]$$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $$1\%,2\%$$ and $$3\%$$ respectively. The maximum percentage error in the detection of the dissipated heat will be :

A screw gauge of pitch $$0.5\mathrm{mm}$$ is used to measure the diameter of uniform wire of length $$6.8\mathrm{cm}$$, the main scale reading is $$1.5\mathrm{mm}$$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is : [Screw gauge has 50 divisions on its circular scale]

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and $$4^{\text{th }}$$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $$1\mathrm{mm}$$. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and $$6^{\text{th }}$$ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :

A torque meter is calibrated to reference standards of mass, length and time each with $$5\%$$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :

An expression of energy density is given by $$u=\frac{\alpha }{\beta }\sin \left(\frac{\alpha x}{kt}\right)$$, where $$\alpha ,\beta$$ are constants, $$x$$ is displacement, $$k$$ is Boltzmann constant and t is the temperature. The dimensions of $$\beta$$ will be :

The dimensions of $$\left(\frac{{\mathrm{B}}^2}{{\mu }_0}\right)$$ will be : (if $${\mu }_0$$ : permeability of free space and $$B$$ : magnetic field)

Consider the efficiency of carnot's engine is given by $$\eta =\frac{\alpha \beta }{\sin \theta }{\log }_e\frac{\beta x}{kT}$$, where $$\alpha$$ and $$\beta$$ are constants. If T is temperature, k is Boltzmann constant, $$\theta$$ is angular displacement and x has the dimensions of length. Then, choose the incorrect option :

Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $$\rho$$ and radius of the drop is r, then $$\mathrm{T}=\mathrm{K}\sqrt{\rho {\mathrm{r}}^3/{\mathrm{S}}^{3/2}}$$ is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.

A travelling microscope has 20 divisions per $$\mathrm{cm}$$ on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

In an experiment to find out the diameter of wire using screw gauge, the following observations were noted : (A) Screw moves $$0.5\mathrm{mm}$$ on main scale in one complete rotation (B) Total divisions on circular scale $$=50$$ (C) Main scale reading is $$2.5\mathrm{mm}$$ (D) $$45^{\text{th }}$$ division of circular scale is in the pitch line (E) Instrument has 0.03 mm negative error Then the diameter of wire is :

Match List I with List II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} List I List II A. Torque I. Nms$${}^{-1}$$ B. Stress II. J kg$${}^{-1}$$ C. Latent Heat III. Nm D. Power IV. Nm$${}^{-2}$$ Choose the correct answer from the options given below :

The speed of light in media 'A' and 'B' are $$2.0\times 10^{10}$$ cm/s and $$1.5\times 10^{10}$$ cm/s respectively. A ray of light enters from the medium B to A at an incident angle '$$\theta$$'. If the ray suffers total internal reflection, then

The refracting angle of a prism is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be -

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $$\stackrel{o}{A}$$ is used to see th object will be :

A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces.

Consider a light ray travelling in air is incident into a medium of refractive index $$\sqrt{2n}$$. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :

A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be : (Given : relative permeability of medium $$\mu$$r = 1)

The difference of speed of light in the two media A and B (vA $$-$$ vB) is 2.6 $$\times$$ 107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c = 3 $$\times$$ 108 ms$$-$$1)