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Physics Physics Ray Optics and Optical Instruments

An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm is placed on other side of lens at 8cm as shown in the figure. Image of object. When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be________ (cm).

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alok kumar singh

Contributor-Level 10

As image coincides with the object, image formed by convex lens should be at the centre of curvature of the convex mirror. If the convex mirror is removed, then, also, image formed the convex lens is at the same location.

Distance of image from object = 12 + 8 + 2 * 15 = 50 cm

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New answer posted

a year ago

Physics Physics Mechanical Properties of Fluids

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is___________________ cm.

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Vishal Baghel

Contributor-Level 10

$p_{1} - p_{2} = \frac{4 T}{r_{1}} . . . . . . . (1)$

$p_{2} - p_{0} = \frac{4 T}{r_{2}} . . . . . . . (2)$

(1) + (2), p₁ – p₀ = 4T $(\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$\Rightarrow \frac{4 T}{r_{e q}} = 4 T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$r_{e g} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{3 * 6}{3 + 6} = 2 c m$

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New answer posted

a year ago

Physics Physics Oscillations

The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum of known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as_______%.

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alok kumar singh

Contributor-Level 10

$L e t, E \propto m^{a} g^{b} T^{c}$

So, $[E] = {[m]}^{a} {[g]}^{b} {[T]}^{c}$

$\Rightarrow M L^{2} T^{- 2} = M^{a} {(L T^{- 2})}^{b} T^{c}$

Thus, E = k mg² T²

$\frac{Δ E}{E} = 2 . \frac{Δ g}{g} + 2 . \frac{Δ T}{T}$

Percentage error in E = 2 * 4 + 2 * 3 = 14%

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a year ago

Physics Physics Wave Optics

White light is passed through a double slit and interference is observed on a screen 1.5m away. The separation between the slits is 0.3mm. The first violet and red fringes are formed 2.0mm and 3.5mm away from the central white fringes. The difference in wavelengths of red and violet light is………nm.

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Vishal Baghel

Contributor-Level 10

For first fringe

$\frac{d y_{1}}{D} = 1 * λ_{1} [n = 1] λ_{1} \to$ Red wavelength

$\frac{d y_{2}}{D} = 1 * λ_{2} [n = 1] λ_{2} \to$ violet wavelength

$\frac{d}{D} (y_{1} - y_{2}) = λ_{1} - λ_{2}$

$y_{1} - y_{2} = \frac{D}{d} (λ_{1} - λ_{2})$

= $\frac{0.3 * 1 0^{- 3}}{1.5} (3.5 - 2) * 1 0^{- 3}$

= 300 * 10^-9

= 300 nm

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a year ago

Physics Physics Moving Charges and Magnetism

If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by proton in a cyclotron to achieve one sixth of the speed of light is_________.

$[m_{p} = 1.67 * 1 0^{- 27} k g, e = 1.6 * 1 0^{- 19} C, S p e e d o f l i g h t = 3 * 1 0^{8} m ? / ? s]$

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alok kumar singh

Contributor-Level 10

2nqV = $\frac{1}{2} m v^{2}$

$n = \frac{m v^{2}}{4 q V}$

$= \frac{1.67 * 1 0^{- 27} * {(0.5 * 1 0^{8})}^{2}}{4 * 1.6 * 1 0^{- 19} * 12 * 1 0^{3}} = 543$

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New answer posted

a year ago

Physics Physics Magnetism and Matter

Two short magnetic dipoles m1 and m2 each having magnetic moment of 1 Am² are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole m2 due to the presence of m₁ is____________ * 10^-7 Nm.

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Vishal Baghel

Contributor-Level 10

$τ_{2} = B_{1} M_{2} S m 90 °$

= B₁ M₂

$= (\frac{μ_{0}}{4 π} * \frac{m_{1}}{r^{3}}) * m_{2}$

= $1 0^{- 7} \frac{1}{1^{3}} * 1$

= 10^-7 N – m

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New answer posted

a year ago

Physics Physics Wave Optics

A source of light is placed in front of a screen. Intensity of light on the screen in I. Two Polaroids P₁ and P₂ are so placed in between the source of light and screen that the intensity of light on screen in $\frac{I}{2} .$ P₂ should be rotated by an angle of_______ (degrees) so that the intensity of light on the screen becomes

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alok kumar singh

Contributor-Level 10

$\frac{l}{2} c o s^{2} ? = \frac{3 l}{8}$

$? c o s ? = \frac{\sqrt[]{3}}{2}$

$? ? = 30 °$

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New answer posted

a year ago

Physics Laws of Motion

The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is_________N.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

$\frac{F_{m a x}}{1 + 2} = 0.5 * 10 \Rightarrow F_{m a x} = 15 N$

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New answer posted

a year ago

Physics System of Particles and Rotational Motion

Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will_________ Mr².

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Vishal Baghel

Contributor-Level 10

= $l_{x x'} = [l_{y y'} + M {(\frac{5 r}{2})}^{2}] + [l_{Z Z'} + M {(\frac{13 r}{2})}^{2}]$

$M (\frac{36 r^{2}}{12}) + \frac{M * 25 r^{2}}{4} = \frac{M r^{2}}{2} + M * \frac{169 M r^{2}}{4}$

= 52 Mr²

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New answer posted

a year ago

Physics Class 11th

The temperature of equals masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. Temperature of mixture when x and z are mixed will be:

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alok kumar singh

Contributor-Level 10

ms_x (16 – 10) = ms_y (20 – 16)

$\Rightarrow \frac{s_{x}}{s_{y}} = \frac{2}{3}$

$m s_{y} (26 - 20) = m s_{z} (30 - 26)$

$\Rightarrow \frac{s_{y}}{s_{z}} = \frac{2}{3}$

$\frac{s_{x}}{s_{z}} = \frac{s_{x}}{s_{y}} * \frac{s_{y}}{s_{z}} = \frac{2}{3} * \frac{2}{3} = \frac{4}{9}$

$m s_{x} (T - 10) = m s_{z} (30 - T)$

->4(T – 10) = 9(30 – T)

->T = 23.84°C

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