An amorphous, transparent and super cooled liquid is called glass.The water glass is the glass obtained by heating the mixture of sodium carbonate or potassium carbonate with silica at 800 degree celsius. This note provides us an information about Glass.
An amorphous, transparent and super cooled liquid is called glass.The water glass is the glass obtained by heating the mixture of sodium carbonate or potassium carbonate with silica at 800 degree celsius. This note provides us an information about Glass.
Things to Remember
An amorphous, transparent and super cooled liquid is called glass.
The water glass is the glass obtained by heating the mixture of sodium carbonate or potassium carbonate with silica at 800 degree celsius.
The ordinary glass is the soft glass obtained after heating the mixture of sodium carbonate, calcium carbonate and silica at high temperature about 1500 degree celsius.
The glass prepared after the heating of potassium carbonate, calcium carbonate and silica at high temperature is hard glass or potash lime glass.
The glass which is manufactured after heating the mixture of sodium carbonate, boron oxide and silica at high temperature is called borosilicate (Pyrex) Glass.
Lead crystal glass is the glass which is obtained after heating the mixture of potassium carbonate, lead monoxide and silica at high temperature.
Multiple colored glass prepared by mixing the raw materials of ordinary glass with different metallic oxides is called colored glass. examples: Cobalt oxide,Iron oxide etc.
MCQs
No MCQs found.
Subjective Questions
Q1:
In the given O is the centre of the circle. If \(\angle QOR = 110°\), calculate the value of the \(\angle PSQ\)
Answer: <p><strong>Solution:</strong></p>
<p>\(\angle QOR = 110°, \angle PSQ= ?\)</p>
<p>\begin{align*} \angle POQ+ \angle QOR &= 180° [\because \text{Sum of 2 adjacent angles}] \\ or, \angle POQ + 110° &= 180°\\ \therefore \angle POQ &= 70°\\ \\ \angle PSQ &= \frac{1}{2}\angle POQ [\because \text{The inscribed angles is half of the centre angle, when both are standing on the same arc.}]\\ \angle PSQ&=\frac{1}{2}\times 70°\\ \therefore \angle PSQ &= 35° \:\: _{Ans} \end{align*}</p>
Q2:
In the given figure, ABCD is a cyclic quadrilateral in which AB is produced to F. IF DC \(\parallel\) BE, \(\angle DAB\)= 90° and \(\angle FBE = 10°\), evaluate the \(\angle ADC \)
Answer: <p>Solution:</p> <p>\(\angle DAB = 92°\\ \angle FBE=10°\\ \angle ADC =?\)</p> <p>\begin{align*} \angle DAB + \angle BCD &= 180° \:\:\: [\because\text {Sum of two opposite angle of cyclic quadrilateral} ] \\ 92°+\angle BCD &= 180°\\ or, \angle BCD &= 180°-92°\\ \therefore \angle BCD &= 88°\\ \angle CBE &= \angle BCD=88° \:\:\:[\because Alternative \: angles ]\\ \angle ADC &= \angle CBF\:\:\: [\because \text{ The exterior angle is equal to the opposite interior angle of cyclic quadrilateral} ] \\ \angle ADC &= 88°+10°\\ \therefore \angle ADC &= 98° \:\: \: _{ans} \end{align*}</p>
Q3:
In the given figure, DCE is a tangent where C is a point of contact. If \(\angle ACB = \angle BDC=x \) and \(\angle BCD = 60°\), find the \(\angle ADC\)
Answer: <p>Solution:</p> <p>\( \angle MON = 140° \\ \angle MPN = ? \)<br>Construction: Take a point A on the circumference of the circle. Join MA and NA.</p> <p>\begin{align*} \angle MAN &= \frac{1}{2}\angle MON \:\:\: [\because \text{The inscribed angle in half of the centre angle when both angle standing on the same arc} ] \\ or, \angle MAN &= \frac{1}{2} \times 140\\ &= 70°\\ Now , \\ \angle MPN + \angle MAN &= 180° [\because \text{The sum of the two opposite angle of the cyclic quadrilateral}]\\ or, \angle MPN + 70° &= 180°\\ or, \angle MPN &= 180°-70°\\ \therefore \angle MPN &= 110°\:\: _{ans} \end{align*}</p>
Q7:
In the given figure, M is the centre of the circle. If \(\angle PQR = b°, \angle RPQ = \frac{3b°}{2}\: and \: \angle RSQ = y°, \) find the value of y.
Answer: <p><strong>Solution:</strong></p> <p>\( \angle RQT = 98°\\ \angle PRQ = \angle RPQ = b° \:\: \: \: \: \: [\because PQ = RQ] \)</p> <p>\begin{align*}\angle RQT &= \angle PRQ + \angle RPQ \:\:\: [\because \text{The exterior angle is equal to the sum of the two opposite interior angles }]\\ b+b&= 98°\\ or, 2b &= 98° \\ or, b &= \frac{98}{2}\\ \therefore b &= 49° \end{align*}</p> <p>\begin{align*} \\ \: \\ \angle QPS + \angle SRQ &= 180° \:\: [\because \text{Sum of the opposite angles of the cyclic quadrilateral. }] \\ b + 44 ° + b+a &= 180° \\ or, 49°+44°+49° + a &= 180°\\ or, a &= 180°-142°\\ \therefore a &= 38° \end{align*}</p>
Q11:
In the given figure, O is the centre of circle and ABCD is a cyclic quadrilateral. BC produce to E. If BC = CD and \(\angle DBC = 33°\), find the value of \(\angle BAD\).
In the adjoining figure, O is the centre of the circle. AC is a diameter. If \(\angle BAC = 2x \: and \: \angle ACB = 3x, \) find the value of \( \angle BDC.\)
In the given figure, C is the centre of the circle. \(\angle DAB =45°, \angle AEB= 60°, \angle CBD=y \: and \: \angle ADB = x,\) find the value of x and y.
In the given figure, O is the centre of the circle. If \(OC \parallel BD, \angle AOC = 70°, \angle OCB = x° \:\: and \:\: \angle CED= y° \) then find the value of x° and y°.
Answer: <p><strong>Solution:</strong></p> <p>\(\angle AOC = 70°\)</p> <p>\begin{align*}\angle AOC &= \angle OBC + \angle OCB \:\: [\because \text{The exterior angle is equal to the two opposite interior angle.}] \\ or, 70° &= x + x \:\:\: [\because OC = OB]\\ or, 2x &= 70°\\ or, x &= \frac{70}{2}\\ \therefore x &= 35° \: \\ \: \\ \angle CBD &= \angle BCO= 35° \:\: [\because Alternative\: angles] \\ \angle OBD &= \angle ODB = 35° + 35° = 70° \:\:\: [\because OB = OD] \end{align*}</p> <p>\( y = \angle EBD + \angle EDB\\ y = 35° + 70° \\ y = 105° \)</p>
Q15:
In the given figure, O is the centre of circle. If \(SO \parallel RQ, \angle POS = 60°, \angle OSQ = x° and \angle STR = y° \) find the value of x and y.
Answer: <p><strong>Solution:</strong></p> <p>\(\angle POS = 60°\)</p> <p>\begin{align*} \angle PQS &= \frac{1}{2} \angle POS \:\: [\because \text{The inscribed angle is half of the centre angle}] \\ &= \frac{1}{2} \times 60\\ &= 30° \\ \: \\ \angle OSQ &= \angle ORQ \:\: [\because \text{Both angle standing on same arc}] \\ x &= 30° \: \: [\because OS = OQ] \\ \angle ORQ &= \angle POS= 60° \:\: [\because OS \parallel QR]\\ \angle SQR &= 60° - 30° = 30° \\ \: \\ \angle STR &= \angle TQR + \angle RQT \:\: [\because \text{The exterior angle is equal to the sum of the two opposite interior angle.}]\\ y &= 60° + 30°\\ \therefore y &= 90° \end{align*}</p>
Q16:
In the given figure O is the centre of a circle. If \(RQ \parallel OP, OR \parallel PQ, \angle ROP = y° \: and \: \angle OPQ = x° \) find the value of x and y.
Answer: <p><strong>Solution:</strong></p> <p>\(\angle VWZ = 58° \\ \angle VOZ = 2 \angle VWZ \: [\because \text{The central angle is double of inscribed angle}] \\ \: \: \: \: x° = 2 \times 58° \\ \therefore x = 116°\\ \: \\ y + 58° = 180° \: [\because \text{ sum of the opposite angles of the cyclic quadrilateral}]\\ y = 180° - 58°\\ \therefore x= 122°\)</p> <p></p>
Videos
Circles: radius, diameter, circumference and Pi | Geometry
Area of a circle
https://www.youtube.com/watch?v=1m9p9iubMLU
Glass
The glass is an amorphous, transparent and supercooled liquid which is prepared from the homogenous mixture of silicates of different metals.
Types of glass
Quartz glass(Silica glass) : The glass which is prepared by heating pure silica at 1600 degree Celsius is called silica glass.It is a pure crystalline glass.
Water glass: The glass which is obtained by heating the mixture of sodium carbonate or potassium carbonate with silica at 800-degree celsius is called water glass. It is called water glass because it is easily soluble in water.
Ordinary glass (Soda glass/Soft glass) : The glass which is obtained after heating the mixture of sodium carbonate, calcium carbonate and silica athigh temperature about 1500 degree celsius is called ordinary glass.It is prepared by heating the mixture of 50% silica, 25% piece of glass, 15% sodiumcarbonate and 10% calcium carbonate.It is called soft glass because it melts at low temperature.
Hard glass (Potash lime glass) : The glass which is prepared after the heating of potassium carbonate, calcium carbonate and silica at high temperature is called hard glass.
Borosilicate (Pyrex) Glass: The glass which is manufacturedafter heating the mixture of sodium carbonate, boronoxide and silica at high temperature is called hard glass.
Lead crystal glass: The glass which is obtained after heating the mixture of potassium carbonate, lead monoxide and silica at high temperature is called lead crystal glass.
Colored glass: The glass of multiple colors which are prepared by mixing the raw materials of ordinary glass with different metallic oxides is called colored glass.
