eSaral - JEE | NEET | Class 8-10 Exam Prep App

How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

Question: How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? Solution: D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain struct...

Match the thermodynamic processes taking place in a system with the correct conditions. In the table :

Question: Match the thermodynamic processes taking place in a system with the correct conditions. In the table : $\Delta Q$ is the heat supplied, $\Delta W$ is the work done and $\Delta U$ is change in internal energy of the system. (I)-(A), (II)-(B), (III)-(D), (IV)-(D)(I)-(B), (II)-(A), (III)-(D), (IV)-(C)(I)-(A), (II)-(A), (III)-(B), (IV)-(C)(I)-(B), (II)-(D), (III)-(A), (IV)-(C)Correct Option: , 4 Solution: (4) (I) Adiabatic process : No exchange of heat takes place with surroundings. $\Righ...

The largest sphere is to be curved out of a right

Question: The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. Solution: The radius of largest sphere $=\frac{h}{2}$ $=\frac{14}{2}$ $=7 \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \times \frac{22}{7} \times(7)^{3}$ $=1437.3 \mathrm{~cm}^{3}$...

In the given figure, the chord AB of the larger of the two concentric circles, with centre O,

Question: In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. Solution: Construction: Join OA, OC and OB We know that the radius and tangent are perperpendular at their point of contact OCA = OCB = 90∘Now, In △OCA and △OCBOCA = OCB = 90∘OA = OB (Radii of the larger circle)OC = OC (Common)By RHS congruency△OCA △OCB CA = CB...

What are the expected products of hydrolysis of lactose?

Question: What are the expected products of hydrolysis of lactose? Solution: Lactose is composed of -D galactose and -D glucose. Thus, on hydrolysis, it gives -D galactose and -D glucose....

A solid is composed of a cylinder with hemispherical ends.

Question: A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Use = 3.1416) Solution: Height of the cylinder $=108-36$ $=72 \mathrm{~cm}$ $r=18 \mathrm{~cm}$ C.S.A. of cylinder $=2 \pi r h$ $=2 \pi \times 18 \times 72$ $=2 \times 3.14 \times 18 \times 72$ $=8138.88 \mathrm{~cm}^{2}$ C.S.A. of 2 hemisphere = surface of a sphere $=4 \p...

Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water

Question: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. Solution: A glucose molecule contains five OH groups while a sucrose molecule contains eight OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water. But cyclohexane and benzene do not contain OH groups. Hence, they cannot undergo H-bonding with water and as a result, are insoluble in water....

In the given figure, PA and PB are the tangents to a circle with centre O.

Question: In the given figure,PAandPBare the tangents to a circle with centreO.Show that the pointsA,O,B,Pare concyclic. Solution: Here, $O A=O B$ $A$ nd $O A \perp A P, O A \perp B P$ (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) $\therefore \angle O A P=90^{\circ}, \angle O B P=90^{\circ}$ $\therefore \angle O A P+\angle O B P=90^{0}+90^{0}=180^{0}$ $\therefore \angle A O B+\angle A P B=180^{\circ}$ (Since, $\angle O A P+\angle O B P+\an...

The height of a solid cylinder is 15 cm

Question: The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm, and height 4 cm are cut off. Find the volume of the remaining solid. Solution: Volume of cylinder $=\pi r^{2} h$ $=\pi \times\left(\frac{7}{2}\right)^{2} \times 15$ $=\frac{735}{4} \pi \mathrm{cm}^{3}$ $=183.75 \pi \mathrm{cm}^{2}$ Volume of cones $=2 \times\left(\frac{1}{3} \pi r^{2} h\right)$ $=24 \pi \mathrm{cm}^{3}$ Volume of remaining solid $=183.75 \pi-24 \pi...

Give plausible explanation for each of the following:

Question: Give plausible explanation for each of the following: (i)Why are amines less acidic than alcohols of comparable molecular masses? (ii)Why do primary amines have higher boiling point than tertiary amines? (iii)Why are aliphatic amines stronger bases than aromatic amines? Solution: (i)Amines undergo protonation to give amide ion. Similarly, alcohol loses a proton to give alkoxide ion. In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on...

Write the reactions of

Question: Write the reactions of(i)aromatic and(ii)aliphatic primary amines with nitrous acid. Solution: (i)Aromatic amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) at 273 278 K to form stable aromatic diazonium salts i.e., NaCl and H2O. (ii)Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2...

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E

Question: In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF Solution: We know that tangent segments to a circle from the same external point are congruent.Now, we haveAD = AF, BD = BE and CE = CFNow, AD + BD = 12 cm .....(1)AF + FC = 10 cm⇒ AD + FC = 10 cm .....(2)BE + EC = 8 cm⇒ BD + FC = 8 cm .....(3)Adding all these we getAD + BD + AD + FC + BD ...

Height of a solid cylinder is 10 cm and diameter 8 cm.

Question: Height of a solid cylinder is 10 cm and diameter 8 cm. Two equal conical hole have been made from its both ends. If the diameter oaf the holes is 6 cm and height 4 cm, find (i) volume of the cylinder, (ii) volume of one conical hole, (iii) volume of the remaining solid. Solution: Height = 10 cm. Radius $=\frac{8}{2}$ $=4 \mathrm{~cm} .$ (i) Volume of cylinder $=\pi r^{2} h$ $=\pi \times(4)^{2} \times 10$ $=160 \pi \mathrm{cm}^{3}$ (ii) Volume of conical hole diameter of cone = 6 cm. ra...

A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K.

Question: A closed vessel contains $0.1$ mole of a monatomic ideal gas at $200 \mathrm{~K}$. If $0.05$ mole of the same gas at $400 \mathrm{~K}$ is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to______ Solution: (266.67) Here work done on gas and heat supplied to the gas are zero. Let $T$ be the final equilibrium temperature of the gas in the vessel. Total internal energy of gases remain same. i.e., $u_{1}+u_{2}=u_{1}^{\prime}+u_{2}^{\prime}$ or, $...

Two concentric circles are of radii 6.5 cm and 2.5 cm.

Question: Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle Solution: We know that the radius and tangent are perperpendular at their point of contactIn right triangle AOPAO2= OP2+ PA2⇒ (6.5)2= (2.5)2+ PA2⇒ PA2= 36⇒ PA = 6 cmSince, the perpendicular drawn from the centre bisect the chord. PA = PB = 6 cmNow, AB = AP + PB = 6 + 6 = 12 cmHence, the length of the chord of the larger circle is 12 cm....

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Question: Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis? Solution: Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide. But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide. Hence, aromatic primary amines cannot be prepared by this process....

Complete the following reactions:

Question: Complete the following reactions: Solution: (i) (ii) (iii) (iv) (v) (vi) (vii)...

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm.

Question: A pointPis 25 cm away from the centre of a circle and the length of tangent drawn fromPto the circle is 24 cm. Find the radius of the circle. Solution: Draw a circle and let $P$ be a point such that $O P=25 \mathrm{~cm}$. Let $T P$ be the tangent, so that $T P=24 \mathrm{~cm}$ Join $O T$, where $O T$ is radius. Now, tangent drawn from an external point is perpendicular to the radius at the point of contact. $\therefore O T \perp P T$ In the right $\triangle O T P$, we have: $O P^{2}=O ...

A circus tent is cylindrical to a height

Question: A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent. Solution: Total canvas used = curved area of cylinder + curved area of cone $=\left(2 \times \frac{22}{7} \times 32.5 \times 3+\frac{22}{7} \times 52.5 \times 53\right) \mathrm{m}^{2}$ $=\frac{22}{7} \times 52.7 \times(6+53) \mathrm{m}^{2}$ $=22 \times 7.5 \times 59...

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’

Question: An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br2and KOH forms a compound C of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C. Solution: It is given that compound C having the molecular formula, C6H7N is formed by heating compound B with Br2and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound B is an amide and compound C is an amine. The only amine having the ...

An iron pillar consists of a cylindrical portion 2.8 m

Question: An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm. Solution: Volume of cylindrical portion $=\pi r^{2} h$ $=\frac{22}{7} \times\left(\frac{20}{2}\right)^{2} \times 280$ $=88000 \mathrm{~cm}^{3}$ Volume of conical portion $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \times \frac{22}{7} \times(10)^{2} \times 42$ $=4400 \mathrm{~cm}^{3}$ Total num...

Give the structures of A, B and C in the following reactions:

Question: Give the structures of A, B and C in the following reactions: (i) (ii) (iii) (iv) (v) (vi) Solution: (i) (ii) (iii) (iv) (v) (vi)...

The specific heat of water

Question: The specific heat of water $=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ and the latent heat of ice $=3.4 \times 10^{5} \mathrm{Jkg}^{-1} \cdot 100$ grams of ice at $0^{\circ} \mathrm{C}$ is placed in $200 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} \mathrm{C}$ is close to (in grams):$61.7$$63.8$$69.3$$64.6$Correct Option: 1 Solution: (1) Here ice melts due to water. Let the amount of ice mel...

A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm.

Question: A pointPis at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn fromPto the circle Solution: Consider the figure.We know that the tangent is perpendicular to the radius of a circle. So, $\mathrm{OPB}$ is a right angled triangle, with $\angle \mathrm{OBP}=90^{\circ}$ By using pythagoras theorem in $\triangle \mathrm{OPB}$, we get $\Rightarrow \mathrm{OB}^{2}+\mathrm{PB}^{2}=\mathrm{OP}^{2}$ $\Rightarrow(20)^{2}+\mathrm{PB}^{2}=(29)^{2}...

The radius of the base of a right circular cone of semi-vertical angle α is r.

Question: The radius of the base of a right circular cone of semi-vertical angle $\alpha$ is $r$. Show that its volume is $\frac{1}{3} \pi r^{3} \cot \alpha$ and curved surface area is $\pi r^{2} \operatorname{cosec} \alpha$. Solution: $\sin \alpha=\frac{r}{l}$ $\Rightarrow r \operatorname{cosec} \alpha=l$ $\tan \alpha=\frac{r}{h}$ $\Rightarrow r \cot \alpha=h$ Volume $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi r^{2} \cdot r \cot \alpha$ $=\frac{1}{3} \pi r^{3} \cot \alpha$ Surface area $=\pi r...

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