Tag: physics

A couple is defined as a pair of two equal and opposite parallel forces acting along two different lines. A couple can produce roatation in the body but not the translational motion. A single force can produce translation motion in the body. Thus the given statement is true that a couple can never be replaced by a single force.

Couple tends to rotate the body but it does not produce translational motion. Thus the given statement is false.

A couple is defined as a pair of two equal and opposite parallel forces acting along two different lines. Since net force acting on the body is zero, so the body is in translatory equilibrium. A couple produces torque which rotates the body. Thus the given statement is true that only a couple can produce pure rotation in a body.

A couple has to be applied to the tap in order to open it. A couple is the combination of two equal and opposite parallel forces acting at different axes. Thus option D is correct.

$\left[ MOI \right] =\left[ M{ R }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ 0 } \right] \ \left[ Pressure \right] =\left[ N/{ M }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] \ \left[ Couple \right] =\left[ N.{ M } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] $

$Power\quad P=100kW\quad =100000W\ w=1800\times \cfrac { 2\pi  }{ 60 } \quad rad/s\ \quad =60\pi \quad rad/s\ P=torque\times w\ torque=530.5\quad Nm$

Total production of cars In year $2004$
$= 70 + 54 + 31 = 155$
Total production of cars in year $2006$
$= 80 + 68 + 38 = 186$
Hence, difference between the total production of three cars in the year $2004$ and $2006$
$= 186- 155 = 31$

Average production of Car $P$
$=$ $\dfrac {\text{Sum of production of Car P in each year}}{\text{Total number of years}}$
$=$ $\dfrac {76 + 82 + 65 + 70 + 85 + 80}{6}$ $=$ $\dfrac {458}{6}$ $= 76.33$
Average production of Car $Q$
$=$ $\dfrac {59 + 62 + 47 + 54 + 57 + 68}{6}$ $=$ $\dfrac {347}{6}$ $= 57.83$
Average production of Car $R$
$=$ $\dfrac {28 + 36 + 42 + 31 + 49 + 38}{6}$ $=$ $\dfrac {224}{6}$ $= 37.33$
Clearly, average production of Car $P$ is greater than Car $Q$ and Car $R$.
Hence, the average production of Car $P$ is maximum.

While considering the motion of an object under the gravitational influence of another object, angular momentum and total mechanical energy are conserved.Linear momentum is not conserved

$\begin{array}{l} F=\dfrac { { G\left( M \right) m } }{ { { { \left( { 3R } \right)  }^{ 2 } } } }  \ \Rightarrow a=\dfrac { { GM } }{ { 9{ R^{ 2 } } } }  \ \therefore \dfrac { { { V^{ 2 } } } }{ { R' } } =a \ \Rightarrow \dfrac { { 2GM } }{ { 3R.R' } } =\dfrac { { GM } }{ { 9{ R^{ 2 } } } }  \ \Rightarrow R'=6R \ Hence, \ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$