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Bunuel
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rohit8865
Bunuel
A circular hole is cut in a circular disc of diameter 10/3 m so that the weight of the disc is reduced to 1/4 of its original weight. The diameter of the hole is

A. \(3*\sqrt 3\) m

B. 4.5 m

C. 5.4 m

D. \(5\sqrt 3\) m

E. \(2*\sqrt 4\) m

area of disc =pi*r^2=pi*25/9
let cutout disc=Pi*r^2
so pi*25/9-Pi*r^2=(1/4)*pi*25/9
solving 2*r(dia) = 4.33~4.5m
Ans B

but i mthinking the dia of cutout disc must be smaller than dia of bigger one which is 10/3 ~3m
but none of option is there less than 3??? :roll:

Plz explain anyone

Thanks

Your approach is similiar to me,but you've made a calculation error on the highlighted portion above

2r=\(\frac{5}{\sqrt{3}}\)
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AbdurRakib
rohit8865
Bunuel
A circular hole is cut in a circular disc of diameter 10/3 m so that the weight of the disc is reduced to 1/4 of its original weight. The diameter of the hole is

A. \(3*\sqrt 3\) m

B. 4.5 m

C. 5.4 m

D. \(5\sqrt 3\) m

E. \(2*\sqrt 4\) m

area of disc =pi*r^2=pi*25/9
let cutout disc=Pi*r^2
so pi*25/9-Pi*r^2=(1/4)*pi*25/9
solving 2*r(dia) = 4.33~4.5m
Ans B

but i mthinking the dia of cutout disc must be smaller than dia of bigger one which is 10/3 ~3m
but none of option is there less than 3??? :roll:

Plz explain anyone

Thanks

Your approach is similiar to me,but you've made a calculation error on the highlighted portion above

2r=\(\frac{5}{\sqrt{3}}\)
might be unnecessary i pulled myself towards any choices above :-D
but yes if any choices is 5/sqrt3 then i have blindly hit that as correct one... :)
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Great question.

I am not sure if I would have enough time to do the calculations in time at the test, however here is my solution.

area of the disc: 25*pi/9
hole: r^2*pi
area left of the disc after erasing the hole = 1/4.

(25*pi/9-r^2*pi)/(25*pi/9)=1/4.
25*pi/9-r^2*pi=25*pi/36
r^2=25/9-25/36
r^2=100/36-25/36
r^2=75/36
r^2= (5*5*3)/(2*3*2*3) = (5*5)/(2*2*3)
r=5/(2*sqrt(3))
d=2r=5/sqrt(3).

As noted before, this does not match any of the answer choices.
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I am also getting the answer as 5/sqrt(3).

Please confirm if I am missing anything or there is a problem with the choices.
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I am confused as to what is 'weight' referring to in this context?
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ameyaprabhu
I am confused as to what is 'weight' referring to in this context?

Weight is simply the weight of the disc. Say 10grams, 20 grams, etc. :-D
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abhimahna
I am also getting the answer as 5/sqrt(3).

Please confirm if I am missing anything or there is a problem with the choices.

Hi

abhimahna Can you please explain this. :think:
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udit harnal
Hi

abhimahna Can you please explain this. :think:

Hi udit harnal ,

Weight will reduce to \(1/4\) means the area will be reduced to \(1/4\) the original area

Or I can say:

Original Area - Removed area = 1/4 Original area --- (1)

Original area = pi \((5/3)^2\)

Removed area = pi \(r^2\)

Substitute these values in (1) above, you will get the answer.

Does that make sense?
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abhimahna Yes got it

Thank You :-D
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