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M grams of steam at

Question: $\mathrm{M}$ grams of steam at $100^{\circ} \mathrm{C}$ is mixed with $200 \mathrm{~g}$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} \mathrm{C}$ [heat of vaporization of water is $540 \mathrm{cal} / \mathrm{g}$ and heat of fusion of ice is $80 \mathrm{cal} / \mathrm{g}]$, the value of $\mathrm{M}$ is_____________ Solution: (40) Using the principal of calorimetry $M_{\mathrm{ice}} L_{f}+m_{\mathrm{ice}}(40-0) C_{w}$ $=m_{\tex...

Why do we require artificial sweetening agents?

Question: Why do we require artificial sweetening agents? Solution: A large number of people are suffering from diseases such as diabetes and obesity. These people cannot take normal sugar i.e., sucrose as it is harmful for them. Therefore, artificial sweetening agents that do not add to the calorie intake of a person are required. Saccharin, aspartame, and alitame are a few examples of artificial sweeteners....

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P.

Question: In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If TPQ = 70∘then TRQ Solution: Construction: Join OQ and OT We know that the radius and tangent are perperpendular at their point of contact∵OTP = OQP = 90∘Now, In quadrilateral OQPTQOT + OTP + OQP + TPQ = 360∘ [Angle sum property of a quadrilateral]⇒ QOT + 90∘+ 90∘+ 70∘= 360∘⇒ 250∘+ QOT = 360∘⇒ QOT = 110∘We know that the angle subtended by an arc at the centre is double the ...

With reference to which classification has the statement,

Question: With reference to which classification has the statement, ranitidine is an antacid been given? Solution: The given statement refers to the classification of pharmacological effects of the drug. This is because any drug that is used to counteract the effects of excess acid in the stomach is called an antacid....

A bucket is in the form of a frustum of a cone and holds 15.25 litres of water.

Question: A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction. Solution: Since volume of frustum $=\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$ $=15250 \mathrm{~cm}^{3}$ $h \times \frac{\pi}{3} \times\left(\left(\frac{25}{2}\right)^{2}+(10)^{2}+\frac{25}{2} \times 10\right)$ $=15250$ $h \times \frac{\pi}{3}(156.25+22.5)$ $=15250$ $h=\frac{3 \t...

Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor,

Question: Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor, Why? Solution: Most drugs when taken in doses higher than recommended may cause harmful effects and sometimes, may even lead to death. Hence, a doctor should always be consulted before taking any medicine....

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures,

Question: Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_{1}$ and $T_{2} .$ The temperature of the hot reservoir of the first engine is $T_{1}$ and the temperature of the cold reservoir of the second engine is $T_{2} . T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$, if both the engines perform equal amount of ...

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

Question: In the given figure, PA and PB are two tangents to the circle with centre O. If APB = 50∘then what is the measure of OAB is Solution: Construction: Join OB We know that the radius and tangent are perperpendular at their point of contact∵OBP = OAP = 90∘Now, In quadrilateral AOBPAOB + OBP + APB + OAP = 360∘ [Angle sum property of a quadrilateral]⇒ AOB + 90∘+ 50∘+ 90∘= 360∘⇒ 230∘+ BOC = 360∘⇒ AOB = 130∘Now, In isoceles triangle AOBAOB + OAB + OBA = 180∘ [Angle sum property of a triangle]⇒...

A frustum of a cone is 9 cm thick and the diameters

Question: A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum.(Take = 22/7). Solution: Volume $=\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$ $=9 \times \frac{\pi}{3}\left(\left(\frac{28}{2}\right)^{2}+\left(\frac{4}{2}\right)^{2}+\frac{28}{2} \times \frac{4}{2}\right)$ $=3 \pi\left((14)^{2}+4+14 \times 2\right)$ $=684 \pi \mathrm{cm}^{3}$ S.A. $=\pi(14+2) \sqrt{(14-2)^{2}}+9=240 \pi \mathrm{cm}^{2}$...

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm,

Question: In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD Solution: We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides. AB + CD = AD + BC⇒6 + 8 = AD + 9⇒ AD = 5 cm...

The radii of the ends of a frustum of a right circular

Question: The radii of the ends of a frustum of a right circular cone are 5 metres and 8 metres and its lateral height is 5 metres. Find the lateral surface and volume of the frustum. Solution: Lateral surface area of frustum $=\pi(\mathbf{r}+\mathbf{R}) \mathbf{l}$ $=\pi(5+8) \times 5$ $=204.28 \mathrm{~m}^{2}$ Height of cone $h=\sqrt{5^{2}-(R-r)^{2}}$ $=\sqrt{5^{2}-(8-5)^{2}}$ $=\sqrt{13}$ Volume $=\frac{\pi \times \sqrt{13}}{3}\left(8^{2}+5^{2}+40\right)=540.56 \mathrm{~cm}^{3}$...

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T.

Question: In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If PBT = 30∘, prove thatBA : AT = 2 : 1 Solution: AB is the chord passing through the centreSo, AB is the diameterSince, angle in a semi circle is a right angleAPB = 90∘By using alternate segment theoremWe have APB = PAT = 30∘Now, in △APBBAP + APB + BAP = 180∘ (Angle sum property of triangle)⇒ BAP = 180∘ 90∘ 30∘= 60∘Now, BAP = APT + PTA (Exterior angle property)⇒ 60∘= 30∘+ P...

What is a biodegradable polymer?

Question: What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. Solution: A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly--hydroxybutyrate-CO-- hydroxyvalerate (PHBV) is a biodegradable aliphatic polyester....

How is dacron obtained from ethylene glycol and terephthalic acid?

Question: How is dacron obtained from ethylene glycol and terephthalic acid? Solution: The condensation polymerisation of ethylene glycol and terephthalic acid leads to the formation of dacron....

The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm.

Question: The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheet required to make this bucket. Solution: Height of the bucket = 30 cm. $r_{1}=21 \mathrm{~cm}$ $r_{2}=7 \mathrm{~cm}$ Therefore, Capacity of the bucket $=\frac{\pi h}{3}\left[r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right]$ $=\frac{22}{7} \times \frac{30}{3}\left[(21)^{2}+21 \times 7+(7)^{2}\right]$ $=20020 \mathrm{~cm}^{3}$ $=20.02$ litres The slant height of the bucket $l=\sqrt{...

Identify the monomer in the following polymeric structures.

Question: Identify the monomer in the following polymeric structures. (i) (ii) Solution: (i) The monomers of the given polymeric structure are decanoic acid $\left[\mathrm{HOOC}-\left(\mathrm{CH}_{2}\right)_{8}-\mathrm{COOH}\right]$ and hexamethylene diamine $\left[\mathrm{H}_{2} \mathrm{~N}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{NH}_{2}\right]$. (ii)The monomers of the given polymeric structure are...

Write the names and structures of the monomers of the following polymers:

Question: Write the names and structures of the monomers of the following polymers: (i)Buna-S(ii)Buna-N (iii)Dacron(iv)Neoprene Solution:...

What are the monomeric repeating units of

Question: What are the monomeric repeating units of Nylon-6 and Nylon-6, 6? Solution: The monomeric repeating unit of nylon 6 is $\left[\mathrm{NH}-\left(\mathrm{CH}_{2}\right)_{5}-\mathrm{CO}\right]$, which is derived from Caprolactam. The monomeric repeating unit of nylon 6,6 is $\left[\mathrm{NH}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}-\mathrm{CO}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CO}\right]$, which is derived from hexamethylene diamine and adipic acid....

Discuss the main purpose of vulcanisation of rubber.

Question: Discuss the main purpose of vulcanisation of rubber. Solution: Natural rubber though useful has some problems associated with its use. These limitations are discussed below: 1.Natural rubber is quite soft and sticky at room temperature. At elevated temperatures ( 335 K), it becomes even softer. At low temperatures ( 283 K), it becomes brittle. Thus, to maintain its elasticity, natural rubber is generally used in the temperature range of 283 K-335 K. 2.It has the capacity to absorb larg...

How does the presence of double bonds

Question: How does the presence of double bonds in rubber molecules influence their structure and reactivity? Solution: Natural rubber is a linear cis-polyisoprene in which the double bonds are present between C2and C3of the isoprene units. Because of this cis-configuration, intermolecular interactions between the various strands of isoprene are quite weak. As a result, various strands in natural rubber are arranged randomly. Hence, it shows elasticity....

Write the name and structure of

Question: Write the name and structure of one of the common initiators used in free radical addition polymerisation. Solution: One common initiator used in free radical addition polymerization is benzoyl peroxide. Its structure is given below....

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC,

Question: In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm,B = 90∘and DS = 5 cm then find the radius of the circle. Solution: We know that tangent segments to a circle from the same external point are congruent.Now, we haveDS = DR, AR = AQNow, AD = 23 cm⇒ AR + RD = 23⇒ AR = 23 RD⇒ AR = 23 5 [∵ DS = DR = 5]⇒ AR = 18 cmAgain, AB = 29 cm⇒ AQ + QB = 29⇒ QB ...

Write the monomers used for getting the following polymers.

Question: Write the monomers used for getting the following polymers. (i)Polyvinyl chloride(ii)Teflon(iii)Bakelite Solution: (i)Vinyl chloride (CH2=CHCl) (ii)Tetrafluoroethylene (CF2= CF2) (iii)Formaldehyde (HCHO) and phenol (C6H5OH)...

The radii of the ends of a bucket of height 24 cm

Question: The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity. (Take = 22/7) Solution: Height of a bucket $=24 \mathrm{~cm}$ $R=15 \mathrm{~cm}$ $r=5 \mathrm{~cm}$ Therefore, Capacity of the bucket $=\frac{\pi h}{3}\left[h^{2}+R r+r^{2}\right]$ $=\frac{22}{7} \times \frac{24}{3} \times\left[(15)^{2}+15 \times 5+(5)^{2}\right]$ $=8171.42 \mathrm{~cm}^{3}$...

Define thermoplastics and thermosetting polymers with two examples of each.

Question: Define thermoplastics and thermosetting polymers with two examples of each. Solution: Thermoplastic polymers are linear (slightly branched) long chain polymers, which can be repeatedly softened and hardened on heating. Hence, they can be modified again and again. Examples include polythene, polystyrene. Thermosetting polymers are cross-linked or heavily branched polymers which get hardened during the molding process. These plastics cannot be softened again on heating. Examples of therm...

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