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Physics Class 12th

A convex lens of focal length 20cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ______cm.

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Payal Gupta

Contributor-Level 10

For refraction of light at lens

$\frac{1}{v} - \frac{1}{- 60} = \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} \Rightarrow \frac{1}{v} = \frac{3 - 1}{60} = \frac{1}{30}$

v = 30 cm

If image has to be formed at object itself then light ray should retrace its path. Hence after refraction at lens, it must strikes normally to the mirror

RM = 20 cm Radius of curvature of mirror

Fm = 10 cm focal length of mirror

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Physics Physics Electrostatic Potential and Capacitance

Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 F. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ________V.

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Payal Gupta

Contributor-Level 10

$\frac{C V + q}{9 C} + \frac{- 3 C V + q}{3 C} = 0$

$\Rightarrow C V + q - 9 C V + 3 q = 0$

$\Rightarrow q = \frac{8 C V}{4} = 2 C V$

$V_{A B} = \frac{C V + q}{9 C} = \frac{3 C V}{9 C} = \frac{V}{3} = \frac{18}{3} = 6 v o l t$

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Physics Physics Waves

Two waves executing simple harmonic motions travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $\sqrt[]{3}$ time amplitude of individual motions. The phase difference between the two motions is _______(degree).

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Payal Gupta

Contributor-Level 10

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Physics Class 12th

Two ideal diodes are connected in the network as shown is figure. The equivalent resistance between A and B is _______ $Ω$ .

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Payal Gupta

Contributor-Level 10

Here diode-D₁ is forced biased and diode-D₂ is reversed biased, so

$R_{A B} = (20 / / 20) + 15 = 25 Ω$

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Physics Class 12th

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ________cm.

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Payal Gupta

Contributor-Level 10

$V = K l \Rightarrow \frac{V_{2}}{V_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{l_{2}}{36} = \frac{1.80}{1.20} = \frac{3}{2} \Rightarrow l_{2} = 54 c m$

$Δ l = l_{2} - l_{1} = 54 - 36 = 18 c m$

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Physics Physics Atoms

$\frac{x}{x + 4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

(i) third permitted energy level to the second level and

(ii) the highest permitted energy level to the second permitted level.

The value of x will be ________.

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Payal Gupta

Contributor-Level 10

According to definition of atomic energy, we can write

$E_{1} = E_{0} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 E_{0}}{36}, a n d E_{2} = E_{0} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$

$\frac{E_{1}}{E_{2}} = \frac{5}{9} \Rightarrow \frac{x}{x + 4} = \frac{5}{5 + 4} \Rightarrow x = 5$

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Physics Physics Thermodynamics

A Block of ice of mass 120g at temperature $0 ° C$ is put in 300g of water at $25 ° C .$ The xg of ice melts as the temperature of the water reaches $0 ° C .$ The value of x is _________.

[Use specific heat capacity of water = 4200Jkg^-1K^-1, Latent heat of ice = 3.5 * 10⁵Jkg^-1]

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Payal Gupta

Contributor-Level 10

Using concept of calorimeter, we can write

Gain in heat = Loss in heat

$x * L = m c Δ T \Rightarrow x * 3.5 * 1 0^{5} = \frac{300}{1000} * 25 * 4200$

$\Rightarrow x = \frac{300 * 25 * 4200}{1000 * 3.5 * 1 0^{5}} = \frac{3 * 25 * 42}{35 * 1000} = \frac{90}{1000} k g = 90 g m$

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Physics Class 11th

Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt[]{x} m .$ The value of x is __________.

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Payal Gupta

Contributor-Level 10

$X_{c} = Y_{c} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{m * 0 + m * 0 + m * 3}{3 m} = 1 m$

$O C = X \sqrt[]{x_{0}^{2} + Y_{c}^{2}} = \sqrt[]{{(1)}^{2} + {(1)}^{2}} = \sqrt[]{2} m$

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Physics Class 11th

A particle is moving in a straight line such that its velocity is increasing at 5 ms^-1 per meter. The acceleration of the particle is _________ms^-2 at a point where its velocity is 20 ms^-1.

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Payal Gupta

Contributor-Level 10

$t a n θ = \frac{d v}{d x} = 5 (m / s) / m$

$\Rightarrow d v = 5 d x \Rightarrow \frac{d v}{d t} = 5 \frac{d x}{d t}$

$\Rightarrow a = 5 v = 5 * 20 = 100 m / s^{2}$

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Physics Class 11th

How to find shift in centre of mass?

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Syed Aquib Ur Rahman

Contributor-Level 10

You should know that when the shift in the centre of mass occurs, you can use the principle of moments. With this logic, we can see the shift from the original body as the combination of two parts. The one that remains part and the other that is removed. The formula for the shift is the product of the mass removed and its distance of its centre of mass from the original centre of mass, which is divided by the mass of the remaining part.

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