s1 Applied Science Labmanual

7/26/2019 s1 Applied Science Labmanual

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(105)

(105)

I . I

. . : I

. : I

. : . A

. : I

.

: .

.PHENOLP HLALEIN: I H 8 9.8.

. : I H 3.1 4.5. .

I .

.

: L 1 2 . I NI N2 N ,

, N1 1 = N2 2

: E .

I , . P

, . B ,

. .

: E . P KM 04) . I

, M . .



. KM 04 .

2 KM O4 + 3 H2 O4 → K2 O4 + 2 M O4 + 3 H20 + 5 0

, . D, KM 04

M . ,

. P , .

. . P .

.

I , ( , )

. A (N M ) .

. , ,

(M ) . .

.

. A :(1) I , (2) I

(3) I , (4) I

.

. . F

HC , H2 O4, KM O4, N OH .

. .....................................



E . N : 1D : ....................

2 3

: 250 0.5 N N2CO3 .:

M N2CO3 = 106

E N2CO3 = M / = 106/2 = 53

N2CO3 1000 1 N N2CO3 = 53

N2CO3 1000 0.5 N N2CO3 = 53/2 = 26.5

N2CO3 250 0.5 N N2CO3 = 26.5/4=6.625

: , , , , .: 6.625 N 2CO3

. 250

. 250 .

.

. F

.

....................................



E . N : 2D : ....................

: 250 0.1 N

M (H2C2O4. 2H20) = 126 E = M /B = 126/2 = 63 1000 1 1N . = 63 1000 0.1 N . = 63/10

6.3 250 0.1N . = 6.3/4

1.575 : , , , .

: 1.575 . O 250 .

. 250 .

.

N , . F

.

.....................................



E . N : 3D : ....................

: (1) N HC1 N OH 4 N OH . (2) HC1 500

. :50 , 20 , 250 .

: HC HC N OH

HC + N OH→ N C + H20

A N OH HC1 . N OH HC

N OH, HC .: A HC .

HC1 , . A ,

C N OH . P 20

N OH . A . . H

N OH , . M .

. .

:

1. N HC = ......................................................... N2. HC 500 = ..................................................

O N OH HC (P )

N B R ( ) HC ( )

I F

= .



N1 1=N2 2

N1= N N OH = ( ) / ( E ) =4/40= 0.1N

1= N OH = 20

2= HC = .N2= N HC =( N1 1 ) / 2 =( 0.1N 20) / ( ) = .NN = ( ) / ( E )( ) = N E

HC 1L , = N HC E HC = N2 36.5 = .. 36.5 = .

HC 500 = /2 = .



E . N : 4D : ....................

2 4

A : (1) N2CO3 5.6 (2) H2 O4 250

. : H2 O4 N 2CO3

N 2CO3 + H2 O4 → 4 N 2 O4 + H20 + CO2

A 2 O4 . M

. F , H2 O4 H2 O4

H2 O4 49. : A N2CO3 . 20

N2CO3 . O

A H2 O4 . A , . H2 O4

. . . F

, H2 O4 .

: 1. N H2

O4

= ....................................N

2. H2 O4 250 =

O N 2CO3 H2 O4 (M )

N B R ( ) H2 O4 ( )

I F

= ..



:N1 1=N2 2

N1= N N2CO3 = ( ) / ( E ) =5.6/53= 0.1056N

1= N2CO3 = 20

2= H2 O4 = .N2= N H2 O4 =( N1 1 ) / 2 =( 0.1056N 20) / ( ) = . NN = ( ) / ( E )( ) = N E

H2 O4 1L , = N H2 O4 E H2 O4 = N2 49 = .. 49 =.

H2 O4 250 = /4 = .



E . N : 5D : ....................

: (1) N OH N2CO3 5.4 H2 O4 (2)

N OH 250 : . F

H2 O4 . N2CO3

N 2CO3 + H2 O4 → N 2 O4 + H20 + CO2

N OH H2 O4. I N OH 2 N OH + H2 O4 → N 2 O4 +

2H20

:

(A) :A

H2 O4 . I H2 O4 , .

20 N2CO3 . O

. I H2 O4 .

. F , 2 O4 .

( ) : A N OH. 20 N OH . O

. I H2 O4 .

. , N OH .

: 1. N N OH = .................................... N2. N OH 250 = .........................

O N 2CO3 H2 O4 (M )

N B R ( ) H2 O4 ( )

I F



= ..

:N1 1=N2 2

N1= N N2CO3 = ( ) / ( E ) =5.4/53= 0.1018N

1= N2CO3 = 202= H2 O4 = .

N2= N H2 O4 =( N1 1 ) / 2 =( 0.1056N 20) / ( ) = . N

O H2 O4 N OH (P )

N B R ( ) H2 O4 ( )

I F

= ..

:N1 1=N2 2

N1

= N H2

O4

.N

1= H2 O4 = .

2= N OH = 20N2= N N OH =( N1 1 ) / 2 =( .N 20) / ( ) = . NN = ( ) / ( E )

N OH 1L , = N N OH E N OH = .. 40 = . N OH 250 = /4 = .



E . N : 6D : ....................

3

: ( ) HNO3

6.2 N OH. (2) HNO3 250

: . F . H2C2O4 + 2N OH →

N 2C2O4 + 2H2O

HNO3 N OH

HNO3 + N OH → N NO3 + H2O

:

1. :A . . A

N OH . 20 N OH . O

N OH , .

2. 3 : A HNO3 . I HNO3 ,

.20

. O . I HNO3

. . . F , HNO3

.:1. N HNO3 = ..................... N

2. HNO3 250 = ..............................

O O N OH ( P )

N B R ( ) O ( )

I F



= ..

:

N1 1=N2 2 N1= N O A = ( ) / ( E ) =6.2/63= 0.0984N

1= O A =

2= N OH =20N2= N N OH =( N1 1 ) / 2 =( . ..) / (20) = N

O N OH HNO3 ( P )

N B R ( ) HNO3 ( )

I F

= ..

:N1 1=N2 2

N1= N N OH .N

1= N OH = 20

2= HNO3 = N2= N HNO3 =( N1 1 ) / 2 =( .N 20) / ( ) = . N

N = ( ) / ( E ) HNO3 1L , = N N OH E N OH = .. 63 = .

N OH 250 = /4 = .



E . N : 7D : .................... 4

A : N KMNO4 500 . 0.1 N F

: P F O4

2 KM O4 + 3 H2 O4 → K2 O4 + 2M O4 + 3H20 + 5 0

2 F O4 + H2 O4 + 0 → F 2( O4)3 + H20 A F O4 KM O4 . F

, KM 04 .

:B , , , .: A KM 04 .

, KM O4 . A ,

.A F O4 . 20 F O4 A 25

.

. F , KM O4 .

: (1) N KM O4 =...........................................N

(2) KM O4 500 = ...........................

O F O4 KM O4

N B R ( ) HC ( )

I F

= ..

: N1 1=N2 2



N1= N F O4 = 0.1N

1= F O4 = 20

2= KM O4 = .N2= N KM O4 =( N1 1 ) / 2 =( 0.1N ..) / ( ) = .NN = ( ) / ( E )

KM O4 1L , = N KM O4 E KM O4 = N2 31.6 = .. 31.6 =.

KM O4 500 = /2 = .



E . N : 8D : .................... 4

: (1) (M ) 39.4

(2) 100 :M , F O4 (NH4)2 O4 . 6 H2O.

I F O4 M . P M

2 KM O4 + 3 H2 O4→ 4 K2 O4 + 2 M O4 + 3 H20 + 5 0

2 F O4 + H2 O4 + 0 → 4 F 2 ( O4)3 + H20

: A M 20 M . A

. A KM O4 . A ,

. KM O4 .

. . F , KM O4 .

: 1. N KM O4 = .................................N2. KM O4 100 = ...................

O M KM O4

N B R ( ) HC ( )

I F

= ..

N1 1=N2 2 N1= N M = 0.1N

1= M = 20

2= KM O4 = .N2= N KM O4 =( N1 1 ) / 2 =( 0.1N ..) / ( ) = .NN = ( ) / ( E )



KM O4 1L , = N KM O4 E KM O4 = N2 31.6 = .. 31.6 =.

KM O4 100 = /10 = .



(105)

: .

: , .: L C = 1 / N . .

= M R+( R LC)

= (R2 2)

: . , . ,

, ,

, . N

.

M R (M R ). R ( R ). N (M R+ R LC).

.,

. R . π (R2 2).

:



1 = .N . = .L = 1 / N . =

=

M =

M D = E R = .

M = ..M = ..

= π (R2 2) = . 3

= . 10 9 3

:

3

Trial no. MSR(mm) VSR MSR + (VSR×LC) Corrected Reading123

Trial no. MSR (mm) VSR MSR + (VSR ×LC)

Corrected Reading

123

Trial no. MSR (mm) VSR MSR + (VSR × LC) Corrected Reading

123



: . : ,

: P . L .

L = / N . H = P R + C H R L = A .

: . F . I

H , . N . B ,

I . . F , 4

( ) . N , .

.

P R (P R). H R. H R H R . N

P R + ( H R LC).

.

. A .



:

D 4 = ..D 1 = ..N . =L = / . = ..

= ..

M = .

A A = . 2

A = .3 = . 10 9 3

:

= 3

Trial no. PSR Observed HSR Corrected HSR PSR+(CorrectedHSR× LC)

123



: . : C ,

: R . .

( ) = 0.01 / (R0 R10 )

R0 R10 10 .

C = + (R R0)

R .

: A ,

. I , .B , (3 2 ).

. R0.B 10 ,

R10 .N



.

R . . N + ( R R0).

( ) = 0.01 / (R0 R10 ) = .. /

C = + (R R0) =

= .. 10 3

: M = . K

Body in

pans

Turning

points

Mean Left Mean Right Resting

pointLeft Right Left Right0 0 R 0 =0 10mg R 10 =Body W=……….g R =



: 1. 2. .

: ,

:A = 4π2 / 2 A 2 .

: 0.6 . (L ).

. . 0.7 , 0.8 , 0.9 1

/2 .

A 2 . A . F

2 4.:

M / 2 =

= 4π 2 / 2 = . / 2 :

1. A = .. /2 2. L = .

Trial no. Length l(m)

Time1

2 Mean t Period T=t/20

T2 l/T 2

1 0.62 0.73 0.84 0.9



: . : H ,

:A H , .H , = M (L / )

.

: . N 50



. 50 . N .

. F , .

:

M (L / )=E =M = M (L / ) = = ..10 3

: M = .

NO. Load (g) Pointer Reading Extension Load/ExtensionLoading Unloading Mean

1

2

3

4

5

0

50

100

150

200



: O . :B , , A , R , K , R

: A , .

/I = R R .:C . 1 . .

0.5 . I , . I .

:

= 1 ). OA P OB Q. OC . OC

(OC 50 ) . P Q

.:

R : M = ..

No. V (volt) I (ampere) R= V/I (ohm)

Sl. No. P Q OA OB OC OC × scale Mean12



1: : M ,

:F ,

.H , AG = M BGO M= ( AG) / BG

: G. B, 5 . N

. (A) . .

AG BG .:

=

M = = 10 3

: M = .

Trial no. BG (cm) AG (cm) M = (W × AG) / BG



2: . : M ,

:A , . I , AO

= GOO , = ( AO) / GO

: G . N , O, GO= 5 .

. , A . AO ,

R GO .( ).

:

M = . = .. 10 3

: M = . K

Trial no. GO (cm) AO (cm) m= (W × AO) / GO



: . :P ,

: I ,

. .

: , 100 (P&Q)

( ). A ( : 50 = 1 ). OA P

OB Q. OC . OC (OC 50 ) .

P Q .:

R : M = ..

Sl. No. P Q OA OB OC OC × scale Mean12



: . : C , ,

: I , = / ( + )

:

. . N /( + ).

.:

: =

Serial no. U V F= uv/(u+v) Mean f



1 : : C ,

: I .

: . D .

:

1 2

M : . 2 :

.

: C , : I ,

.:

. M . . H

.:

1 2Distance between the screen andthe mirror



M R = F =

R : = ..

1 2Distance of the wire gauze fromthe pole

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s1 Applied Science Labmanual