Question 1 |
Match the items in columns I and II.
P-1 Q-4 R-3 S-2 | |
P-1 Q-4 R-2 S-3 | |
P-1 Q-3 R-2 S-4 | |
P-4 Q-1 R-2 S-3 |
Question 2 |
The solution of the differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+ 2xy = e^{-x^{2}}
with y(0)=1
is:
(1+x) e^{+x^{2}} | |
(1+x) e^{-x^{2}} | |
(1-x) e^{+x^{2}} | |
(1-x) e^{-x^{2}} |
Question 2 Explanation:
Given equation
\frac{d y}{d x}+2 x y=\sigma^{x^{2}}
Integrating factor
\begin{aligned} \text { I.F. } &=\mathrm{e}^{\int 2 x \;dx}=e^{x^{2}} \\ \text { Solution is } y x^{e^{x^{2}}}&=\int e^{-x^{2}} \cdot e^{x^{2}} d x \\ \therefore \quad y x^{ e^{x^{2}}}&=x+c\\ d t x &=0 \\ y &=1\\ \therefore \; \; c&=1 \\ y&=(1+x) e^{-x^{2}} \end{aligned}
\frac{d y}{d x}+2 x y=\sigma^{x^{2}}
Integrating factor
\begin{aligned} \text { I.F. } &=\mathrm{e}^{\int 2 x \;dx}=e^{x^{2}} \\ \text { Solution is } y x^{e^{x^{2}}}&=\int e^{-x^{2}} \cdot e^{x^{2}} d x \\ \therefore \quad y x^{ e^{x^{2}}}&=x+c\\ d t x &=0 \\ y &=1\\ \therefore \; \; c&=1 \\ y&=(1+x) e^{-x^{2}} \end{aligned}
Question 3 |
Let x denote a real number. Find out the INCORRECT statement.
S = \left \{ x : x \gt 3 \right \}
represents the set of all real numbers greater than 3 | |
S=\left \{ x : x^{2} \lt 0 \right \}
represents the empty set | |
S=\left \{ x : x \in A \; and \; x \in B \right \}
represents the union of set A and set B | |
S = \left \{ x : a \lt x \lt b \right \}
represents the set of all real numbers between a and b, where a and b are real numbers. |
Question 3 Explanation:
The incorrect statement is, S=\left \{ x : x \in A \; and \; x \in B \right \} represents the union of set A and set B
As and represents the intersection not union.
As and represents the intersection not union.
Question 4 |
A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective.?
\frac{1}{5}
| |
\frac{1}{25} | |
\frac{20}{99} | |
\frac{19}{495} |
Question 4 Explanation:
Probability of first item being defective is
P_{1}=\frac{20}{100}
Probability that second item being defective is
P_{2}=\frac{19}{99}
Probability that both are defective
P=P_{1} P_{2}=\frac{20}{100} \times \frac{19}{99}=\frac{19}{495}
P_{1}=\frac{20}{100}
Probability that second item being defective is
P_{2}=\frac{19}{99}
Probability that both are defective
P=P_{1} P_{2}=\frac{20}{100} \times \frac{19}{99}=\frac{19}{495}
Question 5 |
For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is:
\frac{64T}{\pi d^{3}} | |
\frac{32T}{\pi d^{3}} | |
\frac{16T}{\pi d^{3}} | |
\frac{8T}{\pi d^{3}} |
Question 5 Explanation:
\begin{aligned} \frac{T}{J} &=\frac{\tau}{d / 2} \\ \Rightarrow \qquad \frac{T}{\frac{\pi}{32} d^{4}} &=\frac{\tau}{d / 2} \\ \therefore \qquad & \tau=\frac{16 T}{\pi d^{3}} \end{aligned}
Question 6 |
For a four-bar linkage in toggle position, the value of mechanical advantage is:
0 | |
0.5 | |
1 | |
\infty |
Question 6 Explanation:
\omega_{4} of the output link DC becomes zero at the extreme positions. The extreme positions of the linkage are known as "Toggle position".
\therefore Mechanical advantage =\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
\because \quad \omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
Question 7 |
The differential equation governing the vibrating system is:
m\ddot{x} + c\dot{x} + k(x-y) = 0 | |
m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + kx = 0 | |
m \ddot{x} + c(\dot{x}-\dot{y}) + kx = 0 | |
m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + k(x-y) = 0 |
Question 7 Explanation:
Differential equation governing the above vibration
system is given by
\begin{array}{l} \Rightarrow \frac{m d^{2} x}{d t^{2}}+C\left(\frac{d x}{d t}-\frac{d y}{d t}\right)+k x=0 \\ \Rightarrow m \ddot{x}+c(\dot{x}-\dot{y})+k x=0 \end{array}
Question 8 |
A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. the critical buckling load (P_{cr})
is given by
P_{cr}= \frac{EI}{\pi ^{2}L^{2}} | |
P_{cr}= \frac{\pi ^{2} EI}{3L^{2}} | |
P_{cr}= \frac{\pi EI}{L^{2}} | |
P_{cr}= \frac{\pi^{2} EI}{L^{2}} |
Question 8 Explanation:
P_{E}=\frac{n \pi^{2} E I}{L^{2}}
For pin-ended column
n=1
Here n= End fixity coefficient
E = Modulus of elasticity
I = Second moment of area
L = Actual length of column
For pin-ended column
n=1
Here n= End fixity coefficient
E = Modulus of elasticity
I = Second moment of area
L = Actual length of column
Question 9 |
The number of inversions for a slider crank mechanism is:
6 | |
5 | |
4 | |
3 |
Question 9 Explanation:
No. of links of a slider crank mechanism =4
So there are four inversion of slider crank mechanism.
So there are four inversion of slider crank mechanism.
Question 10 |
For a Newtonian fluid
shear stress is proportional to shear strain | |
rate of shear stress is proportional to shear strain | |
shear stress is proportional to rate of shear strain | |
rate of shear stress is proportional to rate of shear strain |
Question 10 Explanation:
For a Newtonian fluid
\tau \propto \frac{d u}{d y}
\tau=\mu \frac{d u}{d y}
where \quad \tau= shear stress
\frac{d u}{d y}= rate of shear strain
\tau \propto \frac{d u}{d y}
\tau=\mu \frac{d u}{d y}
where \quad \tau= shear stress
\frac{d u}{d y}= rate of shear strain
There are 10 questions to complete.