Physics, asked by ebinfibin, 11 months ago

A body of mass 5kg is taken to the center of earth. what is its mass.?what is its weight there?​

Answers

Answered by ShivamKashyap08
11

Answer:

  • The body's mass (M) will be 5 Kg.
  • The body's weight (W) will be zero.

Given:

  1. Mass of the body (M) = 5 Kg

Explanation:

\rule{300}{1.5}

As we know the Mass is a fundamental quantity which is universally constant. i.e. the value of mass will be constant at every point on this universe.

So, If the Body weighs 5 Kg on the surface of earth then it weighs 5 Kg on the center of the earth.

The body's mass (M) will be 5 Kg at the center of the earth.

\rule{300}{1.5}

\rule{300}{1.5}

From the Formula we know,

\large \bigstar \; \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}

\frak{Here}\begin{cases}\text{$\tt g_d$ Denotes Acceleration due to gravity at depth}\\\text{g Denotes Acceleration due to gravity at surface}\\\text{R Denotes Radius of Earth}\\\text{d Denotes Depth}\end{cases}

Now,

\large \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}

Substituting the values,

\displaystyle \dashrightarrow\tt g_d=g\Bigg(1-\dfrac{d}{R}\Bigg)\\\\\\\dag\;  d=R\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\dfrac{R}{R}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\cancel{\dfrac{R}{R}}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-1\Bigg)\\\\\\\dashrightarrow\tt g_d = g \times 0\\\\\\\dashrightarrow\tt g_d = 0

Multiplying by Mass on both sides.

\displaystyle \dashrightarrow\tt m\times g_d = m\times 0\\\\\\\dashrightarrow\tt W=0\\\\\\\dag\; W=m\;g_d\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt W = 0\; N}}}}

The body's weight (W) will be zero at the center of the earth.

Note:

  • The body is taken to the center of the earth the depth equals to the radius of the earth. So, we took d = R.

\rule{300}{1.5}

Answered by MarshmellowGirl
3

 \large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}

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