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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 21101. |
Why are human sex hormones considered to be lipids? |
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Answer» They are not soluble in water. |
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| 21102. |
अमोनिया से यूरिया बनाने की विधि |
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Answer» 2NH3 + CO2 \(\rightleftharpoons\) NH2COONH4(ammonium carbomate)----------Step(1) NH2COONH4 \(\overset\Delta{\longrightarrow}\) H2O + NH2CONH2(Urea)------ Step(2) ammonia and carbon dioxide are placed into the reactor at high temperature and pressure and the urea is formed in a two steps as given above equations. |
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| 21103. |
write the feature of floating and submerged plants found there |
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Answer» Floating plants have leaves that float on the water surface. Their roots may be attached in the substrate or floating in the water column. Submersed macrophytes are also rooted to the bottom but their leaves grow entirely underwater. |
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| 21104. |
Why is BP of hf greater than hcl |
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Answer» For the given question we can say that the basic reason because of which the Boiling point of HF is greater than HCl is the presence of the greater the intermolecular forces. The greater the boiling point. HF has strong intermolecular forces, due to hydrogen bonding between HF and H. |
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| 21105. |
In a parallel connection what will happen to the light bulbs while the circuit is closed |
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Answer» If the light bulbs connected in parallel, and circuit is closed then electricity will be able to flow first flow burn and other safe. |
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| 21106. |
Remove one bulb what will happen to the other light bulbs will is that so |
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Answer» it depends the lighting might be in series or parallel in series they will switch off because there will be disconnect ie there will be no flow of current |
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| 21107. |
Na series connection what will happen to the light bulbs where the circuit is closed |
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Answer» If the light bulb connected in series, and circuit is closed. then electricity will be able to flow through the bulb, the light bulb glow. |
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| 21108. |
Define electrostatic potential energy? |
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Answer» The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity. |
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| 21109. |
If a curve is represented parametrically by the equations `x=4t^(3)+3` and `y=4+3t^(4)` and `(d^(2)x)/(dy^(2))/((dx)/(dy))^(n)` is constant then the value of n, is |
| Answer» Correct Answer - `16.0` | |
| 21110. |
For the two lines (x - 1)/3 = (y - 2)/-4 = z/2and x = 2 + 3t, y = 4t and z = 2 Identify the correct equations from the following:(A) 4x + 3y − 10 = 0 = 2y + 4z − 4 (B) 4x + 3y + 10 = 0 = 2y + 4z − 4 (C) 4x − 3y − 8 = 0 = z − 2 (D) 4x + 3y + 8 = 0 = z − 2 |
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Answer» Correct option is (A),(C) (A) 4x + 3y − 10 = 0 = 2y + 4z − 4 (C) 4x − 3y − 8 = 0 = z − 2 |
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| 21111. |
The density of a non-uniform rod of length `1m` is given by `rho (x) = a (1 + bx^(2))` where a and b are constants and `0 le x le 1`. The centre of mass of the rod will be atA. `(3(2+b))/(4(3+b))`B. `(4(2+b))/(3(3+b))`C. `(3(3+b))/(3(2+b))`D. `(4(3+b))/(3(2+b))` |
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Answer» Correct Answer - A When `b rarr 0`. The density becomes uniform and hence the centre of mass is at `x = 0.5`. Only option (A) tends to `0.5` as `b rarr 0` |
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| 21112. |
Consider the plane 2x + y + z = 0 and the point P(3, 5, 7). Then, the (A) foot of the perpendicular from P onto the given plane is (−3, 2, 4). (B) image of P in the given plane is (−9, −1, 1). (C) equation of the line joining the points (3, 5, 7) and (−9, −1, 1) is (x - 3)/6 = (y - 5)/6 = (z - 7)/4(D) distance of (−9, −1, 1) from the plane 2x + y + z = 0 is 9. |
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Answer» Correct option is (A),(B),(C) (A) foot of the perpendicular from P onto the given plane is (−3, 2, 4). (B) image of P in the given plane is (−9, −1, 1). (C) equation of the line joining the points (3, 5, 7) and (−9, −1, 1) is (x - 3)/6 = (y - 5)/6 = (z - 7)/4. |
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| 21113. |
Two capacitors `A` and `B` are connected in series with a battery as shown in the figure. When the switch `S` is closed and the two capacitors get charged fully, then A. The potential difference across the plates of A is 4V and across the plates of B is 6VC. The ratio of electric energies stored in A and B is 2 : 3D. The ratio of charges on A and B is `3 : 2` |
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Answer» Correct Answer - B In serues `q_(A) = q_(B) = q = 10 xx (2 xx 3)/(5) xx 10^(-4) = 12 mu C` `V_(A) = (q)/(C_(A)) = 6V " " V_(B) = (q)/(C_(B)) = 4V` In series `U prop (1)/(C ) rArr (U_(A))/(U_(B)) = (C_(B))/(C_(A)) = (3)/(2)` |
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| 21114. |
A `25kg` uniform solid with a `20cm` radius respectively by a verticle wire such then the point suspended is velocity about the center of the sphere torque of `0.10`rad and then montain the sphere then at angle of `1.0` rod minimum the statement if sphere is then roleased its period of the oscillation will beA. `pi` secondB. `sqrt(2) pi` secondC. `2 pi` secondD. `4 pi` second |
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Answer» Correct Answer - D `tau = -k theta` `0.1 = - k (1. 0)`, where k is torsional constant of the wire `k = (1)/(10) T = 2 pi sqrt((1)/(k)) = 2pi sqrt(((2)/(5) xx 25xx (2)^(2))/(1 //10))` `= 2pi sqrt(10 xx 2 xx 2 xx 10) = 4 pi` second |
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| 21115. |
When a hydrogen atom is excited from ground state to first excited state, thenA. its kinetic energy increases by 10.2 eVB. its kinetic energy decreases by 10.2 eVC. its potential energy increases by 20.4 eVD. its angular momentum increases by `1.05 xx 10^(-34) J-s` |
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Answer» Correct Answer - B::C::D ground state `n = 1` first excited state `n = 2` `KE = (1)/(4pi epsilon_(0)) (z = 1)` `KE = (14.4 xx 10^(-10))/(2r) eV` `(KE) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10)) eV = 13.58 eV` `(KE)_(2) = eV = 3.39 eV` KE decreases by `= 10.2 eV` Now `PE = (-1)/(4pi epsilon_(0)) (e^(2))/(r ) = (-14.4 xx 10^(-10))/(r ) eV` `(PE)_(1) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10)) eV = -27.1 eV` `(PE)_(2) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10) xx 4) = -6.79eV` Pe increased by `= 20 4 eV` Now Angular momentum , `L = mvr = (nh)/(2pi)` `L_(2) - L_(1) = (h)/(2pi) = (6.6 xx 10^(-34))/(6.28)` `1.05 xx 10^(-34) J - sec` |
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| 21116. |
How to find TSA, CSA, VOLUME, OUTER CSA, INNER CSA AND EACH BASE AREA OF FOLLOWING 3D SHAPES. I need the formula plz 1.HOLLOW CONE2.HOLLOW HEMISPHERE 3.HOLLOW SPHERE 4.RIGHT CIRCULAR CONE-IN THIS I ONLY WANT THE FORMULA HOW TO FIND EACH base surface area only i do not want rest of my parts for this one only4.hollow cuboid-for this I only want LSA TSA, volume and each base area formula 5.hollow cube - for this i only want LSA, TSA, volume and each base area formula Also plz tell me the difference between hollow cube and solid cube do they have same formula for calculation LSA, TSA and each base surface area and including volume???? Plz ans all these questions fast i am struggling a lot no one is helping me plz ans ?? |
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Answer» 1. Hollow Hemisphere
2.Hollow sphere
3.RIGHT CIRCULAR CONE
4. Hollow cuboid( same as Rectangular Prism)
5. Hollow cube
6. Hollow cone
Yes hollow cube and cuboid have same formula as solid cube and cuboid. The hollow cube is a cube with a cubical hole in the middle.The cube is the Platonic solid (also called the regular hexahedron). It is composed of six square faces that meet each other at right angles and has eight vertices and 12 edges. A cube with unit edge lengths is called a unit cube. This is a cuboid with a rectangular hole, two open, opposite sides and walls of equal thickness. The volume of a cuboid is calculated by multiplying the length by the width and the height of the cuboid. To calculated the volume of a cube, take the edge length of the cube to the power of 3. To determine the volume of a sphere, you have to take the diamater to the power of 3 and multiply it to π as well as 1/6. |
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| 21117. |
Q. How to find base area of a hollow cuboid and solid cuboid. I need the formula to find base area of a hollow cuboid and solid cuboid |
| Answer» Why no ans yet | |
| 21118. |
What is difference betwen hollow cube and solid cube and difference between hollow cuboid and solid cuboid plz ans |
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Answer» Solid Cube: In this, you can imagine a dice. As a dice is purely solid, there no empty space in it. Hollow Cube: In a hollow cube, the middle space is filled with air. It has some thickening. It has two values of length from centre. Same is the case in Solid and Hollow cuboid. |
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| 21119. |
The straight line y = a – x touches the parabola x2 = x – y if a = ....... |
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Answer» line: y = a – x Parabola: y = x – x2 ∴ a – x = x – x2 ∴ x2 – 2x + a = 0 ∵ line touches the parabola, roots should be equal. ∴ B2 – 4AC = (– 2)2 – 4(1)(a) = 0 ∴ 4 = 4a i.e. a = 1. |
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| 21120. |
Let `y gt 0` be the region of space with a uniform and constant magnetic field `B hat(k)`. A particle with charge and mass m travels along the y-axis and enters in magnetic field at origin with speed `v_(0)` in region in particle is subjected to an additional friction force `vec(F)= - k vec(v)`. Assume that particle remains in region `y gt 0`.A. `x=(kmv_(0))/(k^(2)+(qB)^(2))`B. `x=(qBmv_(0))/(k^(2)+(qB)^(2))`C. `y=(kmv_(0))/(k^(2)+(qB)^(2))`D. `y=(qBmv_(0))/(k^(2)+(qB)^(2))` |
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Answer» Correct Answer - B::C `vec(F) = ma = -k (v_(x) hat(i) + v_(y) hat(j)) + q (v_(x)hat(i) + v_(y) hat(j)) xx B hat(k)` `ma_(x) = -kv_(x) + qv_(y) B` `ma_(y) = -kv_(y) - qv_(x)B` At `t = 0, v_(x) = 0" " v_(y) = v_(0) " " x = 0 " " y = 0` finally `v_(x) = 0 " " v_(x) = 0 " " x = x_(1) " " y = y_(1)` `m_(x)o = -kx_(1) - qx_(1) B` `rArr x_(1) = (qBmv_(0))/(k^(2) + (qB)^(2)) rArr y_(1) = (kmv_(0))/(k^(2) + (qB)^(2))` |
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| 21121. |
The shortest distance between the lines (x - 1)/2 = (y - 2)/3 = (z - 3)/4 and (x - 5)/4 = (y - 4)/4 = (z - 5)/5 is (A) 0 (B) √(2/3)(C) 2√3 (D) 3√2 |
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Answer» Correct option is (A) 0 |
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| 21122. |
The symmetric and the parametric forms of the line 2x − y + z − 5 = 0 = x + y − 2z − 1 are (A) (x - 2)/1 = (y + 1)/5 = z/3(B) (x - 2)/-2 = (y + 1)/-10 = z/-6(C) x = 2 + t, y = - 1 + 5t, z = 3t(D) x = 2 + 2t, y = 1 + 5t, z = - 3t |
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Answer» Correct option is (A),(B),(C) (A) (x - 2)/1 = (y + 1)/5 = z/3 (B) (x - 2)/-2 = (y + 1)/-10 = z/-6 (C) x = 2+ t, y = - 1 + 5t, z = 3t |
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| 21123. |
The point of intersection of the lines whose parametric equations are x = 3r + 3, y = − 4r + 2, z = r − 1 and x = − 6t + 1, y = 3t − 2 and z = 1 is (A) (9, 6, −1) (B) (9, −6, 1) (C) (−9, 6, −1) (D) (9, 6, 1) |
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Answer» Correct option is (B) (9, − 6, 1) there is one shortcut too...substitute r = 2 ...then first equation will become x = 9 , y = -6 , z = 1 respectively....and thus the answer |
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| 21124. |
A brick is projected from ground with speed v at angle `theta` from horizontal. The longer face of brick is parallel to ground. The brick slides along ground through some distance after hitting ground and then stops. The collision is perfectly inelastic. The coefficient of friction is `mu(mu lt cot theta)`. The angle `theta` is chosen such that brick travels the maximum horizontal distance before coming to rest. Find the distance in meters : |
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Answer» Correct Answer - A::B Range `= (v^(2) sin 2 theta)/(g)` Impulse along vertical after collision `I_(x) = mv sin theta` Horizontal impulse `I_(y) = - mumv sin theta` So horizontal velocity after impact `= v cos theta - mi v sin theta` So total distance along horizontal `D=(v^(2)2 sin cos theta)/(g)+(v^(2))/(2mu g)(sin theta - mu cos theta)^(2)` `=(v^(2)(4 mu sin theta cos theta+(sin theta-mu cos theta)^(2)))/(2 mu g)` `=(v^(2)(sin theta + mu cos theta)^(2))/(2mug)` `D_(max) = (v^(2)(1+mu^(2)))/(2mu g)`. |
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| 21125. |
There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph : Impulse of deformation is :A. 1 NsB. 3 NsC. 6 NsD. 5 Ns |
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Answer» Correct Answer - B `m_(A)xx0.8 =m_(A)xx0.2 +m_(0)xx1.0` `m_(A)xx0.6 m_(B)xx1.0 m_(B)=0.6 m_(A)` `e=(1-0.2)/(0.8)=1=1.5` `I_(d)=6xx0.5-6xx0=3N-5=10xx(0.8-0.5)=10xx0.3 = 3NS` `Delta U=(1)/(2)xx10xx(0.8)^(2)-(1)/(2)xx16xx(0.5)^(2)=5` `xx0.64 8 xx0.25=3.2 -2.0=1.2J`. |
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| 21126. |
There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph : Coefficient of restitution of the collision isA. 1.5B. 1C. 0.5D. 0.8 |
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Answer» Correct Answer - B `m_(A)xx0.8 =m_(A)xx0.2 +m_(0)xx1.0` `m_(A)xx0.6 m_(B)xx1.0 m_(B)=0.6 m_(A)` `e=(1-0.2)/(0.8)=1=1.5` `I_(d)=6xx0.5-6xx0=3N-5=10xx(0.8-0.5)=10xx0.3 = 3NS` `Delta U=(1)/(2)xx10xx(0.8)^(2)-(1)/(2)xx16xx(0.5)^(2)=5` `xx0.64 8 xx0.25=3.2 -2.0=1.2J`. |
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| 21127. |
There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph : Maximum deformation potential energy is :A. 1.2 JB. 3.2 JC. 2.0 JD. 1.6 J |
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Answer» Correct Answer - A `m_(A)xx0.8 =m_(A)xx0.2 +m_(0)xx1.0` `m_(A)xx0.6 m_(B)xx1.0 m_(B)=0.6 m_(A)` `e=(1-0.2)/(0.8)=1=1.5` `I_(d)=6xx0.5-6xx0=3N-5=10xx(0.8-0.5)=10xx0.3 = 3NS` `Delta U=(1)/(2)xx10xx(0.8)^(2)-(1)/(2)xx16xx(0.5)^(2)=5` `xx0.64 8 xx0.25=3.2 -2.0=1.2J`. |
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| 21128. |
The figure shows a double slit experiment with P and Q as the slits. The path lengths PX and QX are `nlambda` and `(n+2)lambda` respectively, where n is a whole number and `lambda` is the wavelength. Taking the central fringe as zero, what is formed at X? A. First brightB. First darkC. Second brightD. Second dark |
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Answer» Correct Answer - C For brightness, path difference `= n lamda = 2 lamda` So second is bright. |
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| 21129. |
Why do clouds appear white? |
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Answer» Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white. |
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| 21130. |
Statement I: The electrical lamp lights up immediately after swith is onStatement II:The free electrons of wire move with speed of light.Of these statements :(a) Both the statement are true and Statement II is the correct explanation of Statement I.(b) Both the statement are true, but statement II is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true. |
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Answer» (c) Statement I is true, but Statement II is false. |
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| 21131. |
Statement I : Energy is released in nuclear fusion.Statement II : Total binding energy of the product is less than total bindingenergy of reactants in nuclear reaction for nuclear fusion. Of these statements :(a) Both the statement are true and Statement II is the correct explanation of Statement I.(b) Both the statement are true, but statement II is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true. |
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Answer» (a) Both the statement are true and Statement II is the correct explanation of Statement I. |
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| 21132. |
4. Two laser beams one of wave length \( 640 nm \) and the other \( 400 nm \) have same unit flux of photons. Their powers are in the ratioa) \( 64: 40 \)b) \( 1: 1 \)c) \( 5: 8 \)d) \( 25: 64 \) |
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Answer» For same flux energy emitted per unit area per unit time are equal hence their power are equal. 1:1 |
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| 21133. |
Statement I: A large dry cell has higher e.m.f.Statement II: The e.m.f. of a dry cell is proportional to its size.Of these statements :(a) Both the statement are true and Statement II is the correct explanation of Statement I.(b) Both the statement are true, but statement II is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true. |
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Answer» (a) Both the statement are true and Statement II is the correct explanation of Statement I. |
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| 21134. |
In the figure light refraction is shown at the air glass interface when light travels from rare to denser medium,the ray bends towards the normal with angle of refraction less than angle of incidence.(i) lf angle of incidence is iI > 2i, then angle of refraction is(a) 2i(b) >2 i(c) < 2i(d) None of these(ii) For what value of angle of incidence total internal reflection (TIR) takes place?(a) 90º (b) 48º (c) 30º (d) For no value |
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Answer» Correct answer is (i) (c) (ii) (d) |
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| 21135. |
Statement I: Light travels in a straight line in homogeneous medium.Statement II: In diffraction light changes its path without.any change in the medium.(a) Both the statement are true and Statement II is the correct explanation of Statement I.(b) Both the statement are true, but statement II is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true. |
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Answer» (b) Both the statement are true,but statement II is not the correct explanation of Statement I. |
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| 21136. |
The image formed at the retina of human eye is(a) virtual(b) Inverted(c) erect(d) real |
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Answer» Correct answer is (b),(d), |
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| 21137. |
In the figure light refraction is shown at the air-glass interface when light travels from rare to denser medium,the ray bends towards the normal with angle of refraction less than angle of incidence.If angle of incidence is il, > i then angle of refraction is (a) il(b) > il(c) < il(d) None of these |
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Answer» Correct answer is (a) iI |
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| 21138. |
In laser ray, a photon stimulates the emission of another photon. The two photons have(a) same energy(b) same direction(c) same magnitude(d) all of these |
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Answer» Correct answer is (a),(b), |
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| 21139. |
There are two columns.Column I You have to match the correct options of these questions as (a), (b), (c) and (d) from Column II:Column-IColumn-II(i) Frequency of D.C.(a) reduces A.C. current(ii) Choke coil(b) 1/2∈0 E2(iii) Energy density of electic field(c) infinite(iv) 1 kWh(d) 3.6 x 106 J |
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Answer» (i) (c) (ii) (a) (iii) (b) (iv) (d) |
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| 21140. |
Which is/are scalar quantity/ies?(a) Charge(b) Capacitance(c) Potential(d) Electric intensity |
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Answer» Correct answer is (a), (b), (c) |
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| 21141. |
Which of the following oxides is amphoteric is nature ?A. `CaO`B. `CO_(2)`C. `SnO_(2)`D. `SiO_(2)` |
| Answer» Correct Answer - C | |
| 21142. |
which one of the following oxides is neutral?A. `SiO_(2)`B. `CO`C. `ZnO`D. `SnO_(2)` |
| Answer» Correct Answer - B | |
| 21143. |
Due to screening effect of electrons in an atomA. `IE` decreasesB. `IE` increasesC. No change in `IE`D. Attraction of nucleus on the valence electron increases |
| Answer» Correct Answer - A | |
| 21144. |
The order in which the following oxides are arranged according to decreasing basic nature isA. `Na_(2)O, MgO, Al_(2)O_(3), CuO`B. `CuO, Al_(2)O_(3), MgO, Na_(2)O`C. `Al_(2)O_(3), CuO, MgO, Na_(2)O`D. `CuO, MgO, Na_(2), Al_(2)O_(3)` |
| Answer» Correct Answer - A | |
| 21145. |
`I^(-)` or the reaction `I_(2) (g) iff 2I(g)`. `K_(c )=6.25xx10^(-6)` at `900^(@)C`. If 1 mole of `I_(2)` injected into a 4 litre of box at `900^(@)C` initially. Then select correct statement(s). [Given : R = 0.08 L-atom `K^(-1) mol^(-1)`]A. Number of mole of I(g) at equilibrium state = `5 xx 10^(-3)`B. The value of `K_(p)` at `900^(@)C = [(6.25 xx 10^(-6))/(93.84)]` atmC. Equilibrium concentration of I(g) = `12.5 xx 10^(-4)`D. Left moles of `I_(2)(g)` at equilibrium state = 0.9975 |
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Answer» Correct Answer - A::C::D `" "I_(2)(g)iff2l(g)` `"Initial 1 0"` moles `"At equilibrium 1-x 2-x"` `K_(c )=((2x)^(2))/((1-x)^(1))((1)/(V))^(Deltan_(g))` `1-x~=1" "becauseK_(c ) lt 10^(-3)` `K_(c )=(4x^(2))/(1)((1)/(4))^(1)` `x^(2)=K_(c )impliesx=sqrt(K_(c ))=25xx10^(-4)` `[1]=(2x)//4=12.5xx10^(-4)"Molar"` |
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| 21146. |
For similar nature of below graph where `n le 5`, select correct statement `rarr` A. Angular node present in orbital may be 1.B. Angular nodes present in orbital may be 3.C. For possible orbitals magnetic quantum number may be 2.D. For possible orbital value of total nodes must be less than 4. |
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Answer» Correct Answer - A::C n may be 1, 2, 3, 4, 5 n - l - 1 = 2 if l = 1 then n = 4 (possible) if l = 3 then n = 6 (not possible) implies Possible orbitals = 3s / 4p / 5d implies For d orbital m = - 2 to +2 implies Total nodes = n - 1 |
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| 21147. |
The rate consant for the forward reation `A(g) iff 2B(g)` is `1.5 xx 10^(-3)s^(-1)` at 300 K If `10^(-5)` moles of A and 100 moles of B are present in 10L vessel at equilibrium, then select correct statement(s).A. The rate constant of the backward reaction is greater than the rate constant of the forward reaction at given temperature.B. The rate constant of the backward reaction is lesser than the rate constant of the forward reaction at given temperature.C. The value of equilibrium constant `(K_(c ))` is `10^(-8)M`.D. The value of rate constant of the backward direction at given temperature is `1.5 xx 10^(-11) M^(-1)S^(-1)` |
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Answer» Correct Answer - B::D `K=([B]^(2))/([A])=((100//10)^(2))/((10^(-5)//10))=(k_(f))/(k_(b))=(1.5 xx10^(-3))/(k_(b))=10^(8)thereforek_(b)=(1.5xx10^(-3)xx(10^(-6)))/(10^(2))=1.5 xx 10^(-11)` |
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| 21148. |
Classify the following reactions, `CH_(3)underset(CH_(3))underset(|)overset(CH_(3))overset(|)C CH_(2)Br+C_(2)H_(5)bar(O)to` II. `CH_(3)underset(CH_(3))underset(|)overset(CH_(3))overset(|)C CH_(2)Br+C_(2)H_(5)OHto`A. `I:S_(N)1,II:S_(N)2`B. `I:S_(N)2,II:S_(N)1`C. Both `S_(N)1`D. Both `S_(N)2` |
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Answer» Correct Answer - B `C_(2)H_(5)O^(-)` is a strong nucleophie, thus `S_(N)2 C_(2)H_(5) OH` is a weak nucleophle, thus `S_(N)1` |
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| 21149. |
Increasing order of the following alkyl halides for `S_(N)1` reaction is `CH_(3)Cl(I),CH_(3)underset(Cl)underset(|)CHCH_(3)(II),(CH_(3))_(3)C Cl(III)`A. `IltIIltIII`B. `IIltIltIII`C. `IIIltIltII`D. `IltIIIltII` |
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Answer» Correct Answer - A I. Primary II. Secondary III. Tertiary Thus, for `S_(N)1` reaction `I lt II lt III` |
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| 21150. |
Which one of the following C-H bonds is the wekest for homolytic fission ?A. `CH_(3)-H`B. `(C_(6)H_(5))CH_(2)-H`C. `C_(6)H_(5)-H`D. `(C_(6)H_(5))_(3)C-H` |
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Answer» Correct Answer - D `(C_(6)H_(5))_(3)C-H to underset("carbocatinon (most stable)") underset("Reasonance stabilised") ((C_(6)H_(5))C^(o+))` |
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