NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369659
The breaking force for a wire of diameter \(D\) of a material is \(F\).The breaking force for a wire of the same material of diameter \(2 D\) is
1 \(F\)
2 \(2 F\)
3 \(\dfrac{F}{4}\)
4 \(4 F\)
Explanation:
Breaking stress \(=\dfrac{\text { Breaking force }}{\text { Area }}\) \(=\text { Constant }\) \(\therefore \quad \dfrac{F}{\left(\dfrac{\pi D^{2}}{4}\right)}=\dfrac{F^{\prime}}{\pi D^{2}}\) or \(\quad \dfrac{4 F}{\pi D^{2}}=\dfrac{F^{\prime}}{\pi D^{2}}\) \(\Rightarrow \quad F^{\prime}=4 F\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369660
In steel, the Young's modulus and the strain at the breaking point are \(2 \times {10^{11}}\,N{m^{ - 2}}\) and 0.15 respectively. The stress at the break point for steel is
369661
A wire is made of a material of density \(10\;g{\rm{/}}c{m^3}\) and breaking stress \(5 \times {10^9}\;N{\rm{/}}{m^2}\). What length of a wire will break under its own weight when suspended vertically
369662
A lift of mass ' \(M\) ' is tied with thick iron ropes. The maximum acceleration of the lift is ' \(a\) ' \(m/{s^2}\) and maximum safe stress is '\(S'\,N/{m^2}\). The minimum diameter of the rope is
The maximum stress produced in the rope is given by \({\sigma _{\max }} = \frac{{{\text{ maximum force }}}}{{{\text{ area }}}} = \frac{{M(g + a)}}{{\pi {{\left( {\frac{d}{2}} \right)}^2}}} = \frac{{4M(g + a)}}{{\pi d}}\) (when the lift is accelerating upwards with acceleration a) For maximum safe stress, \({\sigma _{\max {\text{ }}}} = S\) \(\begin{aligned}& \Rightarrow S=\dfrac{4 M(g+a)}{\pi d^{2}} \\& \Rightarrow d=\sqrt{\dfrac{4 M(g+a)}{\pi S}}\end{aligned}\)
369659
The breaking force for a wire of diameter \(D\) of a material is \(F\).The breaking force for a wire of the same material of diameter \(2 D\) is
1 \(F\)
2 \(2 F\)
3 \(\dfrac{F}{4}\)
4 \(4 F\)
Explanation:
Breaking stress \(=\dfrac{\text { Breaking force }}{\text { Area }}\) \(=\text { Constant }\) \(\therefore \quad \dfrac{F}{\left(\dfrac{\pi D^{2}}{4}\right)}=\dfrac{F^{\prime}}{\pi D^{2}}\) or \(\quad \dfrac{4 F}{\pi D^{2}}=\dfrac{F^{\prime}}{\pi D^{2}}\) \(\Rightarrow \quad F^{\prime}=4 F\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369660
In steel, the Young's modulus and the strain at the breaking point are \(2 \times {10^{11}}\,N{m^{ - 2}}\) and 0.15 respectively. The stress at the break point for steel is
369661
A wire is made of a material of density \(10\;g{\rm{/}}c{m^3}\) and breaking stress \(5 \times {10^9}\;N{\rm{/}}{m^2}\). What length of a wire will break under its own weight when suspended vertically
369662
A lift of mass ' \(M\) ' is tied with thick iron ropes. The maximum acceleration of the lift is ' \(a\) ' \(m/{s^2}\) and maximum safe stress is '\(S'\,N/{m^2}\). The minimum diameter of the rope is
The maximum stress produced in the rope is given by \({\sigma _{\max }} = \frac{{{\text{ maximum force }}}}{{{\text{ area }}}} = \frac{{M(g + a)}}{{\pi {{\left( {\frac{d}{2}} \right)}^2}}} = \frac{{4M(g + a)}}{{\pi d}}\) (when the lift is accelerating upwards with acceleration a) For maximum safe stress, \({\sigma _{\max {\text{ }}}} = S\) \(\begin{aligned}& \Rightarrow S=\dfrac{4 M(g+a)}{\pi d^{2}} \\& \Rightarrow d=\sqrt{\dfrac{4 M(g+a)}{\pi S}}\end{aligned}\)
369659
The breaking force for a wire of diameter \(D\) of a material is \(F\).The breaking force for a wire of the same material of diameter \(2 D\) is
1 \(F\)
2 \(2 F\)
3 \(\dfrac{F}{4}\)
4 \(4 F\)
Explanation:
Breaking stress \(=\dfrac{\text { Breaking force }}{\text { Area }}\) \(=\text { Constant }\) \(\therefore \quad \dfrac{F}{\left(\dfrac{\pi D^{2}}{4}\right)}=\dfrac{F^{\prime}}{\pi D^{2}}\) or \(\quad \dfrac{4 F}{\pi D^{2}}=\dfrac{F^{\prime}}{\pi D^{2}}\) \(\Rightarrow \quad F^{\prime}=4 F\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369660
In steel, the Young's modulus and the strain at the breaking point are \(2 \times {10^{11}}\,N{m^{ - 2}}\) and 0.15 respectively. The stress at the break point for steel is
369661
A wire is made of a material of density \(10\;g{\rm{/}}c{m^3}\) and breaking stress \(5 \times {10^9}\;N{\rm{/}}{m^2}\). What length of a wire will break under its own weight when suspended vertically
369662
A lift of mass ' \(M\) ' is tied with thick iron ropes. The maximum acceleration of the lift is ' \(a\) ' \(m/{s^2}\) and maximum safe stress is '\(S'\,N/{m^2}\). The minimum diameter of the rope is
The maximum stress produced in the rope is given by \({\sigma _{\max }} = \frac{{{\text{ maximum force }}}}{{{\text{ area }}}} = \frac{{M(g + a)}}{{\pi {{\left( {\frac{d}{2}} \right)}^2}}} = \frac{{4M(g + a)}}{{\pi d}}\) (when the lift is accelerating upwards with acceleration a) For maximum safe stress, \({\sigma _{\max {\text{ }}}} = S\) \(\begin{aligned}& \Rightarrow S=\dfrac{4 M(g+a)}{\pi d^{2}} \\& \Rightarrow d=\sqrt{\dfrac{4 M(g+a)}{\pi S}}\end{aligned}\)
369659
The breaking force for a wire of diameter \(D\) of a material is \(F\).The breaking force for a wire of the same material of diameter \(2 D\) is
1 \(F\)
2 \(2 F\)
3 \(\dfrac{F}{4}\)
4 \(4 F\)
Explanation:
Breaking stress \(=\dfrac{\text { Breaking force }}{\text { Area }}\) \(=\text { Constant }\) \(\therefore \quad \dfrac{F}{\left(\dfrac{\pi D^{2}}{4}\right)}=\dfrac{F^{\prime}}{\pi D^{2}}\) or \(\quad \dfrac{4 F}{\pi D^{2}}=\dfrac{F^{\prime}}{\pi D^{2}}\) \(\Rightarrow \quad F^{\prime}=4 F\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369660
In steel, the Young's modulus and the strain at the breaking point are \(2 \times {10^{11}}\,N{m^{ - 2}}\) and 0.15 respectively. The stress at the break point for steel is
369661
A wire is made of a material of density \(10\;g{\rm{/}}c{m^3}\) and breaking stress \(5 \times {10^9}\;N{\rm{/}}{m^2}\). What length of a wire will break under its own weight when suspended vertically
369662
A lift of mass ' \(M\) ' is tied with thick iron ropes. The maximum acceleration of the lift is ' \(a\) ' \(m/{s^2}\) and maximum safe stress is '\(S'\,N/{m^2}\). The minimum diameter of the rope is
The maximum stress produced in the rope is given by \({\sigma _{\max }} = \frac{{{\text{ maximum force }}}}{{{\text{ area }}}} = \frac{{M(g + a)}}{{\pi {{\left( {\frac{d}{2}} \right)}^2}}} = \frac{{4M(g + a)}}{{\pi d}}\) (when the lift is accelerating upwards with acceleration a) For maximum safe stress, \({\sigma _{\max {\text{ }}}} = S\) \(\begin{aligned}& \Rightarrow S=\dfrac{4 M(g+a)}{\pi d^{2}} \\& \Rightarrow d=\sqrt{\dfrac{4 M(g+a)}{\pi S}}\end{aligned}\)
