ANS:E - 52413

"The hotel was not comfortable."

ANS:A - 5

The sizeof function return the given variable. Example: float a=10; sizeof(a) is 4 bytes Step 1: float arr[] = {12.4, 2.3, 4.5, 6.7}; The variable arr is declared as an floating point array and it is initialized with the values. Step 2: printf('%d\n', sizeof(arr)/sizeof(arr[0])); The variable arr has 4 elements. The size of the float variable is 4 bytes. Hence 4 elements x 4 bytes = 16 bytes sizeof(arr[0]) is 4 bytes Hence 16/4 is 4 bytes Hence the output of the program is '4'.

ANS:D - 3 m

The answer should be A. 1/2m.

With ref of Punmia : the second term in slope correction shall be neglected for slopes flatter than about 1 in 25 (i.e slope should be lesser than 0.04).

The slope is Vertical distance/Horizontal distance.
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Here Vertical distance is h and the horizondal distance is √((L)^2 - (h)^2).

L = 20 given in the problem. We need to find, for which value of h the slope will be less than 0.04.

Sub h = 0.5, we get slope = 0.025;
Sub h = 1, we get slope = 0.050;
Sub h = 2, we get slope = 0.100;
Sub h = 3, we get slope = 0.150; For slopes greater than 5% the above formula must be adopted otherwise the second part of the formula can be neglected.

Since the first three options are less than or equal to 5% they can neglected as the 3m is greater than 5% the second part of formula must included. Hence the answer is (D) 3m. Second term is neglected for slope less than 10%.

Slope for given problem for:

C) (2/20)*100 = 10%.
D) (3/20)*100 = 15%.

So for h = 2 m or less it is neglected.

And for h = 3 m it will be taken into account. For slope greater than 5% closer a closer approximation of corrected slope can be determined by h2/2l + h4/8l3.
Otherwise taken as h2/2l.

In the above example slope taken as 5% of length.
Length = 20 m therefore slope is 1.
Slope = height/length.

We have slope value equal to 1.
Then h2/2l = 1.
Answer is 6.32 m which is more than 3. ** If h is the difference in level between endpoints separated by l, then the slope correction is c=(h^2/2l)+{h^4/8(l^3)}.

** If h is less than 3m in a length often the 20 m, then the quantity h^4/8(l^3) can be neglected. Hence c = h^2/2l.
 

ANS:B - 25 A

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