Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 13 (Q.No. 41)
41.
ABC is rigid bar. It is hinged at A and suspended at B and C by two wires BD and CE made of copper and steel respectively, as shown in the given figure. The bar carries load of 1t at F, mid-way between B and C.
Given Ac = 4 cm2, A = 2 cm2, Ec = 1 x 106 kg/cm2, Es = 2 x 106 kg/cm2.
Ps is the force in the steel wire and Pc is the force in the copper wire, the ratio Pc/Ps will be
Discussion:
9 comments Page 1 of 1.
Karishma said:
3 years ago
AB/AC = 1/2.
(Pc.L/Ac.Ec)/(PS.L/As.Es) = 1/2,
Pc/PS = 1/2.
So, the option is A.
(Pc.L/Ac.Ec)/(PS.L/As.Es) = 1/2,
Pc/PS = 1/2.
So, the option is A.
Fathima Alfina said:
4 years ago
Yes, the Correct answer is 2.
(1)
Fathima Alfina said:
4 years ago
Yes, the Correct answer is 2.
Susmita De said:
4 years ago
Yes, I'm also getting option A.
Since ABC is a rigid bar, according to the theorem of Triangle Deflection at B = Half of the deflection at C.
After solving this, we are getting the right option is A.
Since ABC is a rigid bar, according to the theorem of Triangle Deflection at B = Half of the deflection at C.
After solving this, we are getting the right option is A.
(1)
Ravi kumar said:
4 years ago
The correct answer is A.
Suhaachine said:
5 years ago
Take moment about A.
-BD *1 + 1*1.5 - CE*2 = 0
1.5 = BD + 2 CE
÷1.5
0 = BD/1.5 + 2CE/1.5.
BD/1.5 = 2CE/1.5
So,
BD = 2CE
BD/CE = 2.
So the answer is 2.
Option D.
-BD *1 + 1*1.5 - CE*2 = 0
1.5 = BD + 2 CE
÷1.5
0 = BD/1.5 + 2CE/1.5.
BD/1.5 = 2CE/1.5
So,
BD = 2CE
BD/CE = 2.
So the answer is 2.
Option D.
Jhansi said:
6 years ago
Explain the answer in detail.
Rahul jha said:
8 years ago
The Correct answer is option D.
Dharmendra said:
10 years ago
Please provide the solution.
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