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Physics Mechanical Properties of Solids

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2 months ago

Physics Mechanical Properties of Solids Class 11th

A modulating signal 2sin (6.28 * 10⁶)t is added to the carrier signal 4 sin(12.56 *10⁹)t for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passes through a band pass filter. The bandwidth of the output signal of band pass filter will be_________ MHz.

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Payal Gupta

Contributor-Level 10

Modulating signal → 2sin (6.28 * 10⁶)t

Carrier signal → 4 sin (12.56 * 10⁹)t

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2 months ago

Physics Mechanical Properties of Solids Class 11th

A metal wire of length 0.5 m and cross-sectional area 10^-4 m² has breaking stress 5 * 10⁸ Nm^-2. A block of 10 kg is attached at one of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be ________ ms^-1.

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Payal Gupta

Contributor-Level 10

$l = 0.5 m$

A $= 1 0^{- 4} m^{2}$

Breaking stress, $\frac{F}{A} = 5 * 1 0^{8} N / m^{2}$

$F = 5 * 1 0^{8} * 1 0^{- 4} N$

F = 5 * 104N

T = $\frac{m V^{2}}{l} \Rightarrow V_{(m a x)}^{2} = = \frac{T l}{m}$

$V_{m}^{2} = 2.5 * 1 0^{3}$

Vmax = 50m/s

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2 months ago

Physics Mechanical Properties of Solids Class 11th

Potential energy as a function of r is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$ , where r is the interatomic distance, A and B are positive constant. The equilibrium distance between the two atoms will be

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Payal Gupta

Contributor-Level 10

$\frac{d U}{d r} = 0 \Rightarrow \frac{d}{d r} [\frac{A}{r^{10}}] - \frac{d}{d r} [\frac{B}{r^{5}}]$

$\Rightarrow \frac{d}{d r} [A r^{- 10}] - \frac{d}{d r} [B r^{- 5}] = 0$

$\Rightarrow - 10 A r^{- 11} + 5 B r^{- 6} = 0$

$r = {(\frac{2 A}{B})}^{\frac{1}{5}}$

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2 months ago

Physics Mechanical Properties of Solids Class 11th

The bulk modulus of a liquid is 3 * 10¹⁰ Nm^-2. The pressure required to reduce the volume of liquid by 2% is :

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Vishal Baghel

Contributor-Level 10

We know, $Δ P = - B \frac{Δ v}{v}$ $[? B = - Δ P (\frac{v}{Δ v})]$

$\Rightarrow 100 * Δ P = - B x [\frac{Δ v}{v} * 100]$

$\Rightarrow 100 + Δ P = - 3 * 1 0^{10} (- 2)$

$Δ P = \frac{6 * 1 0^{10}}{100}$

$Δ P = 6 * 1 0^{8} N m^{- 2}$ $\frac{}{}$

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2 months ago

Physics Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A wire of length L is hanging from a fixed support. The length changes to L₁ andL₂when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

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alok kumar singh

Contributor-Level 10

Spring constant of wire (k) = $\frac{A E}{L}$

Case 1

$k [Δ] = 1 g$

$k [L_{1} - L] = 1 g$ - (1)

Case 2 $k [L_{2} - L] = 2 g$ - (2)

$(1) \div (2)$

$\frac{L_{1} - L}{L_{2} - L} = \frac{1}{2}$

L = 2L₁ – L₂

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2 months ago

Physics Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A uniformly heavy rod of mass 20kg, cross sectional area 0.4 m² an length 20m is hanging from a fixe support. Neglecting the lateral contraction. The elongation in the rod due to its own weight is x * 10^-9m. The value of x i________.

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alok kumar singh

Contributor-Level 10

$\int_{0}^{Δ l} d y = \int_{0}^{l} (\frac{m g}{l}) * \frac{d x}{A Y}$

$\Rightarrow Δ l = \frac{m g l}{2 A Y} = 25 * 1 0^{- 9} m$

$\Rightarrow x = 25.00$

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2 months ago

Physics Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The area of cross section of the rope used to lift a load by a crane is 2.5 * 10^-4m². The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be:

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alok kumar singh

Contributor-Level 10

Breading stress = $\frac{W e i g h t}{A r e a}$

$\Rightarrow \frac{10}{2.5 * 10.4} = \frac{25}{A}$

$\Rightarrow A = 6.25 * 1 0^{- 4} m^{2}$

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2 months ago

Physics Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A string of area of cross – section 4 mm² and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produce in the string when the body is at bottom of the circle is________* 10^-5.

(use young's modulus 10¹¹ N/m² and g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

For circular motion:-

$T - m g = \frac{m v^{2}}{r}$

$\Rightarrow T = m (g + \frac{v^{2}}{r})$

$= 2 (10 + \frac{25}{1 / 2}) = 120 N$

$\Rightarrow S t r a i n = \frac{s t r e s}{Y}$

= 30 * 10^-5

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2 months ago

Physics Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The pressure P1 and density d1 of diatomic gas $(γ = \frac{7}{5})$ changes suddenly to P2(>P1) and d2 respectively during an adiabatic process. The temperature of the gas increases and becomes ________times of its initial temperature. (given $\frac{d_{2}}{d_{1}} = 32)$

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3 months ago

Physics Mechanical Properties of Solids Moving Charges and Magnetism

The bulk modulus of a liquid is 3 × 1010 Nm-2. The pressure required to reduce the volume of liquid by 2% is

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