Physics-
General
Easy
Question
A beam of light AO is incident on a glass slab (m = 1.5(D) in the direction shown The reflected ray OB is passed through a Nicol prism On viewing through a Nicol prism, we find on rotating the prism that
- the intensity is reduced down to zero and remains zero
- the intensity reduces down somewhat and rises again
- there is no change in intensity
- the intensity gradually reduces to zero and then again increases
The correct answer is: the intensity gradually reduces to zero and then again increases
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Consider the following reaction and identify (b) 
Consider the following reaction and identify (b)
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Give the decreasing order of the reaction rates of the following benzyl with HBr
Give the decreasing order of the reaction rates of the following benzyl with HBr
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The product (x) and (y) are:
The product (x) and (y) are:
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Coherent light is incident on two fine parallel slits 
and 
as shown in fig If a dark fringe occurs at P, which of the following gives possible phase differences for the light waves arriving at P from and ?
Coherent light is incident on two fine parallel slits and as shown in fig If a dark fringe occurs at P, which of the following gives possible phase differences for the light waves arriving at P from and ?
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There are two paths (I and II) for the preparation of phenyl-2,4-dinitro phenyl ether
Which of the following statements is true?
Path I is feasible, whereas path II is not
Path II is feasible, whereas path I is not
iii. The Cl of (a) undergoes SN reaction because it is activated by the two EWG
groups
iv. The nitration of (c) does not give (b) but it gives because the first nitro group is deactivating, so the second nitro group enters the other ring
There are two paths (I and II) for the preparation of phenyl-2,4-dinitro phenyl ether
Which of the following statements is true?
Path I is feasible, whereas path II is not
Path II is feasible, whereas path I is not
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In the set up shown, the two slits 
are not equidistant from the slit S.
The central fringe at O is then
In the set up shown, the two slits are not equidistant from the slit S.
The central fringe at O is then
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Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
Arrange the following free radicals in the decreasing order of their stability. 
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
Arrange the following free radicals in the decreasing order of their stability.
chemistry-General
chemistry-
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
3, 5–dimethylcyclopentene reacts with N–bromosuccinimide (NBS) in CCl4 in presence of light or peroxide. Identify the product.
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
3, 5–dimethylcyclopentene reacts with N–bromosuccinimide (NBS) in CCl4 in presence of light or peroxide. Identify the product.
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is
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Identify (b) in the following sequence of reactions.
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Major product of the reaction, 
would be
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A hollow sphere of mass M and radius r is immersed in a tank of water (density 
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with the horizontal as shown in the figure. The tension T1 in the wire is :
A hollow sphere of mass M and radius r is immersed in a tank of water (density ). The sphere would float if it were set free. The sphere is tied to the bottom of the tank by two wires which makes angle with the horizontal as shown in the figure. The tension T1 in the wire is :
physics-General
