Question

$\underline{\underline{Question}}$

Please refer to the attachment for the Question.........​

Answer 1

$\huge\underline\bold\pink{Answer}$

Given diameter of the neck is 2 cm
Or we can also consider it as 0.02 m

So let's find area of cross section of the neck firstly

$=>\:\frac{πd^2}{4}$

$=>\:0.000314m^2$

Now , Diameter of the bottom is 10 cm as given in the statement = 0 .1 m

So now , Area of cross section of bottom is

$=>\:\frac{πd^2}{4}$

$=>\:0.00785m^2$

Now , Force applied on cork

Force ( F ) = 1 . 2 Kgf

Let let us assume f as the force acting on bottom therefore ,

$\frac{F}{A}\:=>\:\frac{f}{a}$

$F \: =>\: \frac{f}{a}\:×\:a$

$=>\:\frac{1.2}{0.000314}\:×\:0.00785$

$\huge\green{Hence ,\: we \:get\:30\:Kgf}$

$\huge\underline\bold\red{Thank\:Yuh}$

Answer 2

$\huge{ \underline{ \bold{ᴀɴsᴡᴇʀ....{ \heartsuit}}}}$

Given diameter of the neck is 2 cm

Or we can also consider it as 0.02 m

So let's find area of cross section of the neck firstly

=>\:\frac{πd^2}{4}=>

4

πd

2

=>\:0.000314m^2=>0.000314m

2

Now , Diameter of the bottom is 10 cm as given in the statement = 0 .1 m

So now , Area of cross section of bottom is

=>\:\frac{πd^2}{4}=>

4

πd

2

=>\:0.00785m^2=>0.00785m

2

Now , Force applied on cork

Force ( F ) = 1 . 2 Kgf

Let let us assume f as the force acting on bottom therefore ,

\frac{F}{A}\:=>\:\frac{f}{a}

A

F

=>

a

f

F \: =>\: \frac{f}{a}\:×\:aF=>

a

f

×a

=>\:\frac{1.2}{0.000314}\:×\:0.00785=>

0.000314

1.2

×0.00785