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KVPY QUESTION WITHANSWER(21-10-10)
1.
If from the top of a tower 50 m high, the angles of depression of two objects due north of the tower are respectively 60º and 45º, then the approximate distance between the objects, is :
Sol.
(A) 50 2 2 m
(B) 50 3 3 m
(C) 31 m
(D*) None of these
In triangle triangl e ABC 50 tan 45º = x 50 1= x ....(i) In triangle triangl e ABD
2 g/cm3 3 Also, weight of liquid displaced = Weight of solid body
D=
1 Vdg = VDg 4
3.
Sol.
8 g/cm 3 3 Trimethylamine is a pyramidal pyramidal molecule and formamide is a planar molecule,
x = 50
O || C
xx
N
CH3
H3C
CH3
H
xx
N
H ,
H
1
50 = 50 y 3
The hybridisation of nitrogen in both is
50 + y = 50 3
(A) sp2, sp2.
(B*) sp3, sp2.
y = 50 3 – 50
(C) sp3, sp3.
(D) sp2, sp.
Sol.
y = 50 ( 3 – 1)
2.
1 2 Vdg = V × × g 4 3
d=
50 tan 30º = xy
amide the nitrogen is sp2 hybridised due to tau-
1 A body floats with rd of its volume outside outside 3 3 water and th of its volume outside outside liquid, liquid, 4 then the density of liquid is : 3 8 (A) g/cm3 (B*) g/cm3 8 3 4 9 (C) g/cm3 (D) g/cm3 9 4 If V is the volume of the solid body then, 2 1 Volume of body inside water = 1 V V 3 3 1 3 Volume of body inside liquid = 1 V V 4 4
Weight of water displaced =
2 V×1×g 3
Weight of liquid displaced =
1 V×d×g 4
Where d is density of liquid When a body floats in water then, Weight of the body = weight of water displaced VDg =
In amine, the nitrogen is sp3 hybridised and in
2 V × g (D is the density of solid) s olid) 3
tomerism 4.
A stage of mitosis in which chromosomes get arranged in the form of an equatorial plate in the centre of a dividing cell is called : (A) Prophase (B*) Metaphase (C) Anaphase (D) Telophase