
When a polar bear jumps on to an ice-berg, its weight 200kg wt. is just sufficient to sink the iceberg. Then the weight of iceberg is (specific gravity of ice=0.9; specific gravity of seawater=1.02):
$\begin{align}
& (A)1500kg.wt \\
& (B)1000kg.wt \\
& (C)3000kg.wt \\
& (D)2040kg.wt \\
\end{align}$
Answer
565.5k+ views
Hint:Here we want to find the weight of the ice-berg. For this assume that the upward force and downward forces are equal here. For this first calculate the upward force and then calculate the downward force. On equating the both equations and solving them, we will get the weight of the ice- berg.
Complete answer:
Let the weight of iceberg be $M$ and weight of iceberg is 200kgwt.
We know that,
Density, $\rho =\dfrac{M}{V}$
Then its volume is,
$V=\dfrac{M}{\rho }$
Here Density, $\rho =$ Specific gravity$\times {{10}^{3}}kg/{{m}^{3}}$
Thus by substituting the value of $\rho $ we get,
$\Rightarrow V=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)$
Hence the weight of displaced water$=(V\times 1.02\times {{10}^{3}})$$N$
Net upward force, ${{F}_{upward}}=Mg+200g$
Net downward force, ${{F}_{downward}}=V\rho g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times \left( 1.02\times {{10}^{3}} \right)\times g$
Here, the net upward force and downward forces are equal.
Hence by equating them we get,
$Mg+200g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times \left( 1.02\times {{10}^{3}} \right)\times g$
$\Rightarrow (M+200)\times g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times (1.02\times {{10}^{3}})\times g$
Here both left hand side and right hand side contain $g$ .
Hence cancelling them equation becomes,
$\Rightarrow (M+200)=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times (1.02\times {{10}^{3}})$
$\Rightarrow \left( M+200 \right)=\left( 1.133M \right)$
Taking the terms that containing $M$ in right hand side and other terms in left hand side we get,
$\Rightarrow 200=1.133M-M$
Thus equation becomes,
$\Rightarrow 200=0.133M$
Therefore,
$\Rightarrow M=1500kgwt$
The weight of the iceberg is 1500kgwt.
So, the correct answer is “Option A”.
Additional Information:
Buoyant force is an example of upward force. The material which has less density always flows on water. At that time buoyancy force will be equal to force of gravity.
Note:
The force that exerts on the surface of the object is called the normal force. For an object at rest, the net force on that object will be equal to zero. In that case the upward force and the downward force will be equal. Here, the downward force is the weight exerted by the object and upward force is the normal force.
Complete answer:
Let the weight of iceberg be $M$ and weight of iceberg is 200kgwt.
We know that,
Density, $\rho =\dfrac{M}{V}$
Then its volume is,
$V=\dfrac{M}{\rho }$
Here Density, $\rho =$ Specific gravity$\times {{10}^{3}}kg/{{m}^{3}}$
Thus by substituting the value of $\rho $ we get,
$\Rightarrow V=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)$
Hence the weight of displaced water$=(V\times 1.02\times {{10}^{3}})$$N$
Net upward force, ${{F}_{upward}}=Mg+200g$
Net downward force, ${{F}_{downward}}=V\rho g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times \left( 1.02\times {{10}^{3}} \right)\times g$
Here, the net upward force and downward forces are equal.
Hence by equating them we get,
$Mg+200g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times \left( 1.02\times {{10}^{3}} \right)\times g$
$\Rightarrow (M+200)\times g=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times (1.02\times {{10}^{3}})\times g$
Here both left hand side and right hand side contain $g$ .
Hence cancelling them equation becomes,
$\Rightarrow (M+200)=\left( \dfrac{M}{0.9\times {{10}^{3}}} \right)\times (1.02\times {{10}^{3}})$
$\Rightarrow \left( M+200 \right)=\left( 1.133M \right)$
Taking the terms that containing $M$ in right hand side and other terms in left hand side we get,
$\Rightarrow 200=1.133M-M$
Thus equation becomes,
$\Rightarrow 200=0.133M$
Therefore,
$\Rightarrow M=1500kgwt$
The weight of the iceberg is 1500kgwt.
So, the correct answer is “Option A”.
Additional Information:
Buoyant force is an example of upward force. The material which has less density always flows on water. At that time buoyancy force will be equal to force of gravity.
Note:
The force that exerts on the surface of the object is called the normal force. For an object at rest, the net force on that object will be equal to zero. In that case the upward force and the downward force will be equal. Here, the downward force is the weight exerted by the object and upward force is the normal force.
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

