1 Atomic Structure
If atoms of elements have completely filled outermost shell then their combining capacity or valency is : 1) one 2) two 3) zero 4) five
Solution
If atoms of elements, have a completely filled outermost shell then their combining capacity or valency is zero.
2 Ionic Equilibrium
The compound whose 0.1 M solution is basic is
1) Sodium acetate
2) Ammonium sulphate
3) Ammonium chloride
4) Ammonium acetate
Sodium acetate is a salt of weak acid acetic acid and strong base sodium hydroxide. For aqueous solution of sodium acetate, pH=7+0.5pKa?+0.5logc. Hence, pH will always grater than 7 and the solution will be basic.
3 Classification of Elements and Periodicity
The atoms/ion listed in correct order of increasing size are
1) Be2+, Mg2+, Na+
2) Na+, Cl -, K+
3) Al, Na, S
4) Na, Si, H
4 Sets relations and functions
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, f(x) = 9x4+12x3 -36x2 +25, x Î R , then :
1) S1 = {-2,1}; S2 ={0} 2) S1 = {-2}; S2 ={0, 1} 3) S1 = {-2,1}; S2 ={1} 4) S1 = {-1}; S2 ={0, 2}
SOLUTION
5 Differential Equations
The degree and order of the differential equation of the family of all parabolas whose axis is x-axis are respectively-
1) 2, 1 2) 1, 2 3) 3, 2 4) 2, 3
Equation is y2 = 4a(x – h)
differentiate we get
2y dy/dx = 4a
Again differentiate then
yd2y/dx2 + (dy/dx)2 =0
Clearly degree 1 and order 2
6 Definite Integrals
Find
7 P – Block Elements
Which one of the following acid possesses oxidising, reducing and complex forming properties?
1) HNO3 2) HC1 3) H2SO4 4) HNO2
8 Work Power Energy
A body of mass 1kg begins to move under the action of a time dependent force F=(2ti+3t2 k)N. What will be power developed by the force
1) (2t2+3t3 ) W 2) (2t2+4t4 ) W 3) (2t3+3t4 ) W 4) (2t3+3t5 ) W
SOLUTION:−
9 Basic Principles of Organic Chemistry
Tautomerismwill be exhibited by 1) (CH3)3CNO 2) (CH3)2NH 3) R3CNO2 4) RCH2NO2
only RCH2NO2 has α-hydrogen atoms - > will shows tautomerism
10 Probability
In a hurdle race, a runner has probability $$p$$ of jumping over a specific hurdle. Given that in $$5$$ trials, the rummer succeeded $$3$$ times, the conditional probability that the runner had succeeded in the first trial is
1) $$\dfrac {3}{5}$$ 2) $$\dfrac {2}{5}$$ 3) $$\dfrac {1}{5}$$ 4) $$\dfrac {4}{5}$$
Let $$A$$ denote the event that the runner succeeds exactly $$3$$ times out of five and $$B$$ denote the event that the number succeeds on the first trial. $$\displaystyle P(B|A)=\frac {P(B\cap A)}{P(A)}$$ But $$P(B\cap A)=P$$ (succeeding in the first trial and exactly once in two other trials) $$=p(^4C_2p^2(1-p)^2)=6p^3(1-p)^2$$ and $$P(A)=^5C_3p^3(1-p)^2=10p^3(1-p)^2$$ Thus, $$\displaystyle P(B|A)=\frac {6p^3(1-p)^2}{10p^3(1-p)^2}=\frac {3}{5}$$.
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