Tag: forces and matter

Questions Related to forces and matter

A cable that can support a load of 800 N is cut into two equal parts. The maximum load that can be supported by either part is 

  1. 100 N

  2. 400 N

  3. 800 N

  4. 1600 N


Correct Option: C
Explanation:

Breaking stress $= \dfrac{800 N}{A} \Rightarrow BA$

  $F = 800 N$
Breaking stress doesn't depend upon the length of the cable.

A uniform steel bar of cross-sectional area A and length L. is suspended so that it hangs vertically. The stress at the middle point of the bar is ( $\rho $ is the density of steel)

  1. $\frac{L}{2A} \rho g$

  2. $\frac{L\rho g}{2} $

  3. $\frac{LA}{\rho g}$

  4. $L\rho g$


Correct Option: C

On suspending a weight $Mg$ the length $l$ of elastic wire and area of cross section $A$ its length becomes double the initial length. The instantaneous stress action on the wire is:

  1. $\dfrac{Mg}{A}$

  2. $\dfrac{Mg}{2A}$

  3. $\dfrac{2Mg}{A}$

  4. $\dfrac{4Mg}{A}$


Correct Option: C

The velocity of the transverse waves in a wire of density $8000kg/m^3$ is $300 m/s$. The tensile stress in the wire is then

  1. $7.2\times10^8 N/m^2$

  2. $6.8\times10^8 N/m^2$

  3. $5.2\times10^8 N/m^2$

  4. $8.4\times10^8 N/m^2$


Correct Option: A
Explanation:

Given density = 8000 $kg/{ m }^{ 3 }$ and velocity =300 m/s

Also we know velocity of transverse wave  $\text{V}=\sqrt { \dfrac { T }{ \mu  }  } \text{where T= tension and}\quad \mu =\text{mass per unit lenght}\quad =\dfrac { m }{ L } \quad $
also $\text{Density}\quad \rho =\dfrac { mass }{ Area\times Lenght } \ \rho \times A=\dfrac { m }{ L } $
$\text{Tension T}=\rho A{ V }^{ 2 }$
We know stress $\sigma =\dfrac { Force }{ Area } =\dfrac { Tension }{ Area } =\dfrac { { V }^{ 2 }\rho A }{ A } ={ V }^{ 2 }\rho $
${ 300 }^{ 2 }\times 8000=7.2\times { 10 }^{ 8 }\dfrac { N }{ { m }^{ 2 } } $ 

An external force of $10\ N$ acts normally on a square area of each side $50\ cm$. The stress produced in equilibrium state is

  1. $10\ N/m^{2}$

  2. $20\ N/m^{2}$

  3. $40\ N/m^{2}$

  4. $50\ N/m^{2}$


Correct Option: C

The length of wire is increased by $0.06\%$ by a load of $40N$ whose tensile modulus is $20\times10^{10}N/M^2$.The subjected stress is 

  1. $12\times10^{10}N/m^2$

  2. $1.2\times10^{8}N/m^2$

  3. $120N/m^2$

  4. $1.25\times10^6N/m^2$


Correct Option: B
Explanation:

$\cfrac{\triangle l}{l}\times 100=0.06$

$\implies \cfrac{\triangle l}{l}=\cfrac{0.06}{100}$
Now stress $=\cfrac{20\times 10^{10}\times 0.06}{1000}\=1.2\times 10^8\ N /m^2$

According to $C.E$ van der Waal, the interatomic potential varies with the average interatomic distance $(R)$ as 

  1. $R^{-1}$

  2. $R^{-2}$

  3. $R^{-4}$

  4. $R^{-6}$


Correct Option: D
Explanation:

According to the relation

$ V(r)=\dfrac{-3}{4}\dfrac{{{\alpha }^{2}} _{0}l}{{{(4\pi {{\varepsilon } _{0}})}^{2}}{{R}^{6}}} $

$ V(r)\propto \dfrac{1}{{{R}^{6}}} $

$ V(r)\propto {{R}^{-6}} $


Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2\% $ and $0.002\%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is $\left( {Y = 2.0 \times {{10}^{11}}N{m^{ - 2}}} \right)$

  1. $3.2 \times {10^9}N{m^{ - 2}}$

  2. $3.2 \times {10^7}N{m^{ - 2}}$

  3. $3.6 \times {10^9}N{m^{ - 2}}$

  4. $3.6 \times {10^7}N{m^{ - 2}}$


Correct Option: C

A steel wire has an ultimate strength of above $2.0 \times 10 ^ { 7 } \mathrm { kg } - \mathrm { w } \mathrm { J } / \mathrm { m } ^ { 2 }$ . How large a load can a
0.7$\mathrm { cm }$ in diameter steel wire hold before breaking?

  1. $700 \mathrm { kg } - \mathrm { wt }$

  2. $770 \mathrm { kg } - \mathrm { wt }$

  3. $300 \mathrm { kg } - \mathrm { wt }$

  4. None


Correct Option: A

If equal and opposite forces applied to a body tend to elongate it, the stress so produced is called

  1. Tensile stress

  2. Compressive stress

  3. Tangential stress

  4. Working stress


Correct Option: A