Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ such that $\mathrm{d}_{2}=2 \mathrm{~d}_{1}$ and $l>\mathrm{d}_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_{2}=2 \theta_{1}$. If the shear moduli of material 1 is $4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, then shear moduli of material 2 is $\mathrm{x} \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, where value of x is _______ .
Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ such that $\mathrm{d}_{2}=2 \mathrm{~d}_{1}$ and $l>\mathrm{d}_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_{2}=2 \theta_{1}$. If the shear moduli of material 1 is $4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, then shear moduli of material 2 is $\mathrm{x} \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, where value of x is _______ .
Show Hint
Correct Answer: 1
Approach Solution - 1
Approach Solution -2
Let's carefully analyze the problem and verify step-by-step why the correct answer is \( x = 1 \).
Given Data:
- Two slabs of materials (1) and (2)
- Thicknesses: \( d_2 = 2d_1 \)
- Equal shearing force \( F \) applied
- Angles of deformation: \( \theta_2 = 2\theta_1 \)
- Shear modulus of material 1: \( G_1 = 4 \times 10^9 \, \mathrm{N/m^2} \)
- We need: \( G_2 = x \times 10^9 \, \mathrm{N/m^2} \)
Concept Used:
The shearing strain is given by \( \theta = \frac{x}{d} \), and for equal shearing force and area, the shear stress \( \tau = \frac{F}{A} \) is the same for both materials.
From the definition of shear modulus:
\[ G = \frac{\text{shear stress}}{\text{shear strain}} = \frac{\tau}{\theta} \] \[ \Rightarrow G \propto \frac{1}{\theta} \]
However, since the slabs have different thicknesses, the strain also depends on \( d \). Using geometry of deformation,
\[ \theta = \frac{x}{d} \Rightarrow x = \theta d \]
For equal force, the lateral displacement \( x \) is proportional to \( \frac{1}{G} \) (from \( \tau = G\theta \)). Combining, we get:
\[ \theta \propto \frac{d}{G} \] \[ \Rightarrow \frac{\theta_2}{\theta_1} = \frac{d_2 / G_2}{d_1 / G_1} = \frac{d_2 G_1}{d_1 G_2} \]
Step-by-Step Solution:
Step 1: Substitute the given ratio \( \frac{\theta_2}{\theta_1} = 2 \) and \( d_2 = 2d_1 \):
\[ 2 = \frac{2 G_1}{G_2} \]
Step 2: Simplify the expression:
\[ G_2 = G_1 \] \[ G_2 = 4 \times 10^9 \, \mathrm{N/m^2} \]
Step 3: Therefore, the value of \( x = 4/4 = 1 \).
Final Computation & Result:
The shear modulus of material (2) is:
\[ \boxed{G_2 = 1 \times 10^9 \, \mathrm{N/m^2}} \]
Final Answer: \( x = 1 \)
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Elasticity Questions
- Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A) : The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is Elasticity.
Reason (R) : The restoring force depends upon the bonded inter atomic and inter molecular force of solid.
In the light of the above statements, choose the correct answer from the options given below : - With rise in temperature, the Young's modulus of elasticity
- The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is :
- A wire of cross-sectional area $A$, modulus of elasticity $2 \times 10^{11} \, \text{Nm}^{-2}$, and length $2 \, \text{m}$ is stretched between two vertical rigid supports.
When a mass of $2 \, \text{kg}$ is suspended at the middle, it sags lower from its original position making angle $\theta = \frac{1}{100}$ radian on the points of support.
The value of $A$ is ____ $\times 10^{-4} \, \text{m}^2$ (consider $x \ll l$).
(Given: $g = 10 \, \text{m/s}^2$)
- One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of the upper wire to that of the lower wire will be:\[\text{[Area of cross section of wire = 0.005 cm}^2, \, Y = 2 \times 10^{11} \, \text{N/m}^2, \, g = 10 \, \text{m/s}^2\text{]}\]
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.