2015 (December) CSIR Part A (Solution): Engineering Science, Chemical Science and physical Science.

1.Each of the following pairs of words hides a number, based on which you can arrange them in ascending order. Pick the correct answer:

I. Cloth reel

J. Silent Wonder

K. Good tone

L . Bronze rod

• L,K,J,I
• I,J,K,L
• K,L,J,I
• K,J,I,L

Solution:

I. Cloth reel

J. Silent Wonder

K. Good tone

L . Bronze rod

So, Correct order is : L,K,J,I

1st option is correct.

2. Which of the following values is same as \(2^{2^{2^2}}\).

• \(2^6\)
• \(2^8\)
• \(2^{16}\)
• \(2^{222}\)

Solution: \(2^{2^{2^2}}\)=\(2^16\).

3.The Maximum number of points formed by intersection of all pairs of diagonals of convex octagon is

• 70
• 400
• 120
• 190

Solution: For intersection of two diagonal we require \(4\) points on the polygon. So, Ways in which we can select \(4\) vertices out of \(8\)= \(\binom{8}{4}\)=\(70\), 1st option is correct.

4. Find the height of the box of base area 24 cm X 48 cm, in which the longest stick can be kept is 56 cm long?

• 8 cm
• 32 cm
• 37.5 cm
• 16 cm

Solution: Let height =\(h\).

  

So , we have \(56^2=h^2+24^2+48^2\) \(\Rightarrow h^2=56^2-(24^2+48^2)\)\(\Rightarrow h=16\).

4th option is correct.

5. An infinite row of boxes is arranged, Each box has half the volume of the previous box. If the largest box has volume of 20 cc, what is the total volume of the boxes?

• infinite
• 400 cc
• 40 cc
• 80 cc

Solution:  Sequence of volume of boxes :

\(20,10,5,5/2,.... \)

Sum of infinite terms of G.P.= \(\frac{a}{1-r}\) for \(r<1\) Here \(r=1/2\) So Sum=\(\frac{20}{0.5}\)=\(40\)cc. Third option is correct.

6. A \(12\) m \(\times \) \(4\) m rectangular roof is resting on four \(4\) m tall thin poles. Sunlight falls on the roof at an angle of \(45^o \) From the east, creating a Shadow on the ground, what will be the area of the shadow?

• \(24 \quad m^2\)
• \(36 \quad m^2\)
• \(48\quad m^2\)
• \(60\quad m^2\)

Solution:

Area will remain the same.

7. Let \(r\) be a positive number satisfying \(r^{\frac{1}{1234}}+r^{\frac{-1}{1234}}=2\) then \(r^{4321}+r^{-4321}\)=?

• 2
• \(2^\frac{4321}{1234}\)
• \(2^{3087}\)
• \(2^{1234}\)

Solution: \(r^{\frac{1}{1234}}+r^{\frac{-1}{1234}}=2\)

Squaring both sides we get \({r^{\frac{1}{1234}}}^2+{r^{\frac{-1}{1234}}}^2+2=4\)

\(\Rightarrow r^{\frac{1}{617}}+r^{\frac{-1}{617}}=2\)

Repeating the same process, we get, \(r+\frac{1}{r}=2\) Again squaring many times we get

\(r^{4321}+r^{-4321}=2\)

8. Find the missing term:

Solution:

Mirror image along x axis then rotation through \(180^0\). So answer will be 2.

9. A man starts his journey at 0100 hrs local time to reach another country at 0900 Hrs local time on the same date. He starts a return journey on the same night at 2100 Hrs local time to his original place, taking the same time to travel back. If the time zone of his country visit lags by 10 Hrs, the duration for which the man was away from his place is

• 48 Hrs
• 20 Hrs
• 25Hrs
• 36Hrs

Solution:

So total time=\(18+18+12=48\)Hrs

1st option is correct.

10: If you change only one observation from a set of 10 observations, which of the following will definitely change?

• Mean
• Median
• Mode
• Standard Deviation

Solution: Mean=\(\sum x/n\). Mean Depends on all the observations. So Mean will definitely change.

11. Consider a series of letters placed in the following way:

U_G_C_C_S_I_R

Each letter moves one step to its right and the extreme right letter takes the first position, completing one operation. After which of the following numbers of operations do the C's not sit side by side?

• 3
• 19
• 10
• 25

Solution: After 7 such operations arrangement will be the same.

So,  after \(7+4=11\) operations. C's will not be together.

S0, 2nd option is correct.

12.what is the maximum number of parallel, non-overlapping cricket pitches(length 24 m, width 3 m ) that can be laid in a field of diameter 140 m , if the boundary is required to be at least 60 m from the center of any pitch ?

• 6
• 7
• 12
• 4

Solution:

Now Pitch of width \(3\)m in \(20\) m \(=6+1\) (Since boundary in measured from the center of pitch.

So answer is \(7\).

13. If

Here a,b,c,d are digits then a+b=

• 4
• 9
• 11
• 16

Solution: since last digit of \(a\times 2=6\). Therfore, \(a=3\) or \(8\).

If \(a=3\)then, \(c=4\) and then since last digit of \(b\times 3\) is 4. So, \(b=8\). And then last step is not possible. So, \(a\neq 3\).

So, \(a=8\). and then we have;So \(a+b=11\)

14. ABC is a right angled triangle inscribed in a semicircle, Smaller semicircles are drawn on the sides BC and AC. If the area of the triangle is a, what is the total area of the shaded region?

Solution:

Area of region shaded yellow= \(\pi \times \frac{{\frac{\sqrt{h^2+b^2}}{2}}^2}{2}-\frac{1}{2}bh\)

=\(\frac{1}{2} \times (\pi\times \frac{h^2+b^2}{4}-bh)\)

=\(\frac{1}{8}(\pi(h^2+b^2)-4\times b\times h\)

Therefore area of shaded region =\(\frac{1}{2}(\pi {\frac{b}{2}}^2+{\frac{h}{2}}^2)-\frac{1}{8}(\pi(h^2+b^2)-4\times b\times h\)=\(\frac{1}{2}\times b\times h\)=\(a\).

15.An ant can lift another ant of its size whereas an elephant cannot lift another elephant of its size, because

• ant muscle fibers are stronger than elephant muscle fibers.
• ant has proportionally thicker legs than elephant
• strength scales as the squares of the size while weight scales as cube of the size.
• ants work cooperatively, whereas elephants work as individuals.

Solution: 3rd option is correct. \(S \propto M^{2/3}\)

See : http://hep.ucsb.edu/courses/ph6b_99/0111299sci-scaling.html

16. An inclined plane rests against a horizontal cylinder of radius R. If the plane makes an angle \(30^o\) with the ground, the point of contact of the plane with cylinder is at a height of

• 1.500 R
• 1.866 R
• 1.414 R
• 1.000 R

Solution:

From picture height = \(R+\sin 60^o R =1.8666 R\)

17. The product of the perimeter of a triangle, the radius of its in-circle, and a number gives the area of the triangle. The number is:

• \(\frac{1}{4}\)
• \(\frac{1}{3}\)
• \(\frac{1}{2}\)
• 1