Which one of the following will be the major product when is tr-Turito

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Easy

Chemistry-

Which one of the following will be the major product when

Is treated with dilute H₂SO₄ in the presence of HgSO₄

Chemistry-General

Answer:The correct answer is:

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The appropriate reagent (s) for the transformation
is / are

Transformation of C = O group into CH₂ in the presence of Zn + Hg is called Clemmensen’s reduction and in the presence of NH₂ - NH₂/OH^- is called Wolf-Kishner Reduction

The appropriate reagent (s) for the transformation
is / are

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Transformation of C = O group into CH₂ in the presence of Zn + Hg is called Clemmensen’s reduction and in the presence of NH₂ - NH₂/OH^- is called Wolf-Kishner Reduction

For which of the following molecule significant m ¹ 0
I)
II)
III)
IV)

-OH and -SH groups do not cancel their dipole moment

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II)
III)
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-OH and -SH groups do not cancel their dipole moment

Y is

Y is

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The most suitable reagent of r the conversion of RCH₂OH → RCHO is

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X and Y are respectively

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Which is correct about the change given below

Which is correct about the change given below

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Which alkali metal floats over cold water without any reaction?

4, Na, k all are less denses them water among these first group Li is very less reactive

Which alkali metal floats over cold water without any reaction?

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4, Na, k all are less denses them water among these first group Li is very less reactive

For the given uniform square lamina $A B C D$ , whose centre is $O$

Let the each side of square lamina is

d

I subscript E F end subscript equals I subscript G H end subscript

(due to symmetry)
And

I subscript A C end subscript equals I subscript B D end subscript

(due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript

I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript

(ii)
From Eqs. (i) and (ii), we get

I subscript A C end subscript equals I subscript E F end subscript

therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction

equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses

I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

For the given uniform square lamina $A B C D$ , whose centre is $O$

physics-General

Let the each side of square lamina is

d

I subscript E F end subscript equals I subscript G H end subscript

(due to symmetry)
And

I subscript A C end subscript equals I subscript B D end subscript

(due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript

I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript

(ii)
From Eqs. (i) and (ii), we get

I subscript A C end subscript equals I subscript E F end subscript

therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction

equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses

I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

Moment of inertia of the system about the centre of plane is given by

I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

equals 1.004 plus 2.008 plus 3.012 plus 4.016

equals 10.04 blank k g minus m to the power of 2 end exponent

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

physics-General

Moment of inertia of the system about the centre of plane is given by

I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

equals 1.004 plus 2.008 plus 3.012 plus 4.016

equals 10.04 blank k g minus m to the power of 2 end exponent

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

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I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

blank equals fraction numerator 1 over denominator 2 end fraction.4.2

equals 4 blank

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

blank equals fraction numerator 1 over denominator 2 end fraction.4.2

equals 4 blank

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is